diff --git a/resources_servers/physics_judge/app.py b/resources_servers/physics_judge/app.py index 85ab2b9d08b..8af9d7f3574 100644 --- a/resources_servers/physics_judge/app.py +++ b/resources_servers/physics_judge/app.py @@ -34,7 +34,8 @@ bidirectional A/B swap done by ``math_with_judge`` is skipped. 4. Per-domain breakdown via ``compute_subset_metrics(field="domain")`` so the Tier-2 domain-stratified accuracy that Skills exposes through - ``subset_for_metrics=domain`` is preserved. + ``subset_for_metrics=domain`` is preserved. ``verify()`` carries each + row's ``domain`` into its response, since the parent's models drop it. Skills places ``{generation}`` in the judge prompt; we re-use that placeholder name when filling the YAML so the prompt is character-for-character identical. @@ -58,6 +59,8 @@ JudgeEvaluation, LibraryJudgeMathResourcesServer, LibraryJudgeMathResourcesServerConfig, + LibraryJudgeMathVerifyRequest, + LibraryJudgeMathVerifyResponse, ) @@ -86,6 +89,14 @@ class PhysicsJudgeResourcesServerConfig(LibraryJudgeMathResourcesServerConfig): ) +class PhysicsJudgeVerifyRequest(LibraryJudgeMathVerifyRequest): + domain: Optional[str] = None + + +class PhysicsJudgeVerifyResponse(LibraryJudgeMathVerifyResponse): + domain: Optional[str] = None + + class PhysicsJudgeResourcesServer(LibraryJudgeMathResourcesServer): """math_with_judge subclass with NeMo Skills' physics judge.""" @@ -107,6 +118,12 @@ def model_post_init(self, context: Any) -> None: super().model_post_init(context) self._judge_prompt_config: PromptConfig = load_prompt_config(self.config.judge_prompt_path) + async def verify(self, body: PhysicsJudgeVerifyRequest) -> PhysicsJudgeVerifyResponse: + # The parent's models have no `domain`; carry it into the verify row, which is where + # compute_metrics reads it for the per-domain breakdown. + response = await super().verify(body) + return PhysicsJudgeVerifyResponse(**response.model_dump(), domain=body.domain) + async def _verify_answer( self, question: str, expected_answer: str, generated_answer: str ) -> tuple[float, Optional[str], float, Optional[list[JudgeEvaluation]]]: diff --git a/resources_servers/physics_judge/data/example_rollouts.jsonl b/resources_servers/physics_judge/data/example_rollouts.jsonl index 94a2b2ad695..532dc4fb0ce 100644 --- a/resources_servers/physics_judge/data/example_rollouts.jsonl +++ b/resources_servers/physics_judge/data/example_rollouts.jsonl @@ -1,5 +1,5 @@ -{"responses_create_params":{"background":null,"include":null,"input":[{"content":"Below is an open-ended problem in Physics. Please answer this problem adhering to the following rules:\n1. Please use LaTeX format to represent the variables and formulas used in the solution process and results.\n2. Please put the final answer(s) in \\boxed{}, note that the unit of the answer should not be included in \\boxed{}.\n3. If there are multiple final answers, please seperated them by commas in \\boxed{}, e.g., \\boxed{answer 1, answer 2}.\nProblem: What is the de Broglie wavelength of a particle with momentum $p$?","role":"user","type":"message"}],"instructions":null,"max_output_tokens":null,"max_tool_calls":null,"metadata":null,"model":null,"parallel_tool_calls":true,"previous_response_id":null,"prompt":null,"reasoning":null,"service_tier":null,"store":null,"temperature":null,"text":null,"tool_choice":"auto","tools":[],"top_logprobs":null,"top_p":null,"truncation":null,"user":null,"stream":null},"response":{"id":"resp_rWPwxggSyIMlON6Fp_EtQmrbpnV0OTYcVDJF0Sw6U6ygsjR4YRFRqgGWTbjpvL77rRISP6rrdeLMM9bFhgZw8t2LX24bT6otNpw03wYrL4VBRwjtoJhjCa6MC5WfidCNN3RV2AjSe8ICjD5V2JUNd3X4CXfvc_k21xA5Vvr3_DOYDgBhrE3bP9hn-m6OjxOLtgp1Punw7LXVC8T5t8vI29WW3q30LLtfKP5XcKucXJBgqwKL9kY7RO9VCqiod7yDfGxdcJIJl8_SgQOjYpS6GDLYrVQLi3D0VaQIQD2zrWu1fzdmnwpy2ok83zRoX--s7r62j2Ny38JbJmrmYzzJuMJnUmPzWg==","created_at":1777570222.0,"error":null,"incomplete_details":null,"instructions":null,"metadata":{},"model":"gcp/google/gemini-2.5-pro","object":"response","output":[{"id":"rpHzabyKD4-Pq8YPvqPZoAU","content":[{"annotations":[],"text":"According to the de Broglie hypothesis, all matter exhibits wave-like properties. The wavelength associated with a particle, known as the de Broglie wavelength, is related to its momentum.\n\nThe de Broglie wavelength, denoted by $\\lambda$, is given by the formula:\n$\\lambda = \\frac{h}{p}$\n\nwhere:\n- $\\lambda$ is the de Broglie wavelength.\n- $h$ is Planck's constant, a fundamental constant in physics with a value of approximately $h \\approx 6.626 \\times 10^{-34} \\text{ J} \\cdot \\text{s}$.\n- $p$ is the momentum of the particle.\n\nThus, the de Broglie wavelength of a particle with momentum $p$ is the ratio of Planck's constant to the momentum.\n\n\\boxed{\\frac{h}{p}}","type":"output_text","logprobs":null}],"role":"assistant","status":"completed","type":"message"}],"parallel_tool_calls":false,"temperature":0.0,"tool_choice":"auto","tools":[],"top_p":null,"background":null,"conversation":null,"max_output_tokens":null,"max_tool_calls":null,"previous_response_id":null,"prompt":null,"prompt_cache_key":null,"reasoning":null,"safety_identifier":null,"service_tier":null,"status":"completed","text":{"format":null,"verbosity":null},"top_logprobs":null,"truncation":null,"usage":{"input_tokens":125,"input_tokens_details":{"cached_tokens":0,"audio_tokens":null,"text_tokens":125},"output_tokens":1163,"output_tokens_details":{"reasoning_tokens":980,"text_tokens":183},"total_tokens":1288,"cost":null},"user":null,"store":null,"provider_specific_fields":{"traffic_type":"ON_DEMAND"}},"reward":1.0,"expected_answer":"\\boxed{h/p}","extracted_answer":"h/p","library_reward":1.0,"judge_evaluations":null,"_ng_task_index":3,"_ng_rollout_index":0,"agent_ref":{"type":"responses_api_agents","name":"physics_judge_simple_agent"}} -{"responses_create_params":{"background":null,"include":null,"input":[{"content":"Below is an open-ended problem in Physics. Please answer this problem adhering to the following rules:\n1. Please use LaTeX format to represent the variables and formulas used in the solution process and results.\n2. Please put the final answer(s) in \\boxed{}, note that the unit of the answer should not be included in \\boxed{}.\n3. If there are multiple final answers, please seperated them by commas in \\boxed{}, e.g., \\boxed{answer 1, answer 2}.\nProblem: A block of mass $m$ slides down a frictionless inclined plane of angle $\\theta$. What is its acceleration along the slope?","role":"user","type":"message"}],"instructions":null,"max_output_tokens":null,"max_tool_calls":null,"metadata":null,"model":null,"parallel_tool_calls":true,"previous_response_id":null,"prompt":null,"reasoning":null,"service_tier":null,"store":null,"temperature":null,"text":null,"tool_choice":"auto","tools":[],"top_logprobs":null,"top_p":null,"truncation":null,"user":null,"stream":null},"response":{"id":"resp_TEuGNod0Jgxvlowu1BMv-KlWV1pQEG7mZ0kde2eb8elfBYN9jV4yXNKMs8ZbS3GUPhkCu5Hyk0Ybkot1KY2j88IvvV_UpaaQ1IQWxRNpEQL7ImxaVK1sQ6Z9ui9lfCSXjhEE47wRFdZnT1GOtZEYp6UXv1Stb6Rda-SLxeApBetRoH_4c0O87CtqkftMUifTFPyikbGeealqQ8Ha6GHTStyT2GKDPqaUZx9ITWueKpsdxDiq8JucuLN9EOLIT4Rf-U1WRTe7R4UDu65pWne9xjA5oL0TRJP5J6Mv_Qr1_5r65YdrNiFH5ghmzdcEgvVmlzyjCAwqnQF-4wv7U7cc-3Om19PJKA==","created_at":1777570222.0,"error":null,"incomplete_details":null,"instructions":null,"metadata":{},"model":"gcp/google/gemini-2.5-pro","object":"response","output":[{"id":"rpHzabunEMuepucPsd6p6Ac","content":[{"annotations":[],"text":"Here is a step-by-step solution to the problem.\n\nLet the mass of the block be $m$ and the angle of the inclined plane be $\\theta$. The acceleration due to gravity is denoted by $g$.\n\n1. **Identify the forces acting on the block.**\n There are two forces acting on the block:\n * The gravitational force, $F_g = mg$, acting vertically downwards.\n * The normal force, $N$, exerted by the inclined plane on the block, acting perpendicular to the surface of the plane.\n Since the plane is frictionless, there is no friction force.\n\n2. **Choose a suitable coordinate system.**\n It is convenient to choose a tilted coordinate system where the x-axis is parallel to the inclined plane (pointing down the slope) and the y-axis is perpendicular to the inclined plane (pointing upwards).\n\n3. **Resolve the forces into components.**\n We resolve the gravitational force, $F_g$, into components parallel and perpendicular to the inclined plane.\n * The component of the gravitational force parallel to the slope is $F_{g,x} = mg \\sin(\\theta)$. This component pulls the block down the slope.\n * The component of the gravitational force perpendicular to the slope is $F_{g,y} = mg \\cos(\\theta)$. This component pushes the block against the plane.\n The normal force $N$ acts entirely along the positive y-axis.\n\n4. **Apply Newton's Second Law.**\n Newton's Second Law, $\\Sigma \\vec{F} = m\\vec{a}$, is applied along each axis of the coordinate system. Let $a$ be the acceleration of the block along the slope (in the x-direction).\n\n * **Along the y-axis (perpendicular to the slope):**\n The block does not accelerate in the direction perpendicular to the plane, so the net force in the y-direction is zero.\n $\\Sigma F_y = N - mg \\cos(\\theta) = 0$\n This gives $N = mg \\cos(\\theta)$.\n\n * **Along the x-axis (parallel to the slope):**\n The only force component acting along the x-axis is the parallel component of gravity.\n $\\Sigma F_x = mg \\sin(\\theta)$\n According to Newton's Second Law, the net force in the x-direction is also given by $\\Sigma F_x = ma$.\n Therefore, we can equate the two expressions:\n $ma = mg \\sin(\\theta)$\n\n5. **Solve for the acceleration.**\n To find the acceleration $a$, we can cancel the mass $m$ from both sides of the equation:\n $a = g \\sin(\\theta)$\n\nThis is the acceleration of the block sliding down the frictionless inclined plane.\n\nThe final answer is:\n\\boxed{g \\sin(\\theta)}","type":"output_text","logprobs":null}],"role":"assistant","status":"completed","type":"message"}],"parallel_tool_calls":false,"temperature":0.0,"tool_choice":"auto","tools":[],"top_p":null,"background":null,"conversation":null,"max_output_tokens":null,"max_tool_calls":null,"previous_response_id":null,"prompt":null,"prompt_cache_key":null,"reasoning":null,"safety_identifier":null,"service_tier":null,"status":"completed","text":{"format":null,"verbosity":null},"top_logprobs":null,"truncation":null,"usage":{"input_tokens":136,"input_tokens_details":{"cached_tokens":0,"audio_tokens":null,"text_tokens":136},"output_tokens":2149,"output_tokens_details":{"reasoning_tokens":1516,"text_tokens":633},"total_tokens":2285,"cost":null},"user":null,"store":null,"provider_specific_fields":{"traffic_type":"ON_DEMAND"}},"reward":1.0,"expected_answer":"\\boxed{g\\sin\\theta}","extracted_answer":"g*sin(theta)","library_reward":1.0,"judge_evaluations":null,"_ng_task_index":0,"_ng_rollout_index":0,"agent_ref":{"type":"responses_api_agents","name":"physics_judge_simple_agent"}} -{"responses_create_params":{"background":null,"include":null,"input":[{"content":"Below is an open-ended problem in Physics. Please answer this problem adhering to the following rules:\n1. Please use LaTeX format to represent the variables and formulas used in the solution process and results.\n2. Please put the final answer(s) in \\boxed{}, note that the unit of the answer should not be included in \\boxed{}.\n3. If there are multiple final answers, please seperated them by commas in \\boxed{}, e.g., \\boxed{answer 1, answer 2}.\nProblem: A spaceship moves at speed $v$ relative to the Earth. Express the time dilation factor $\\gamma$ in terms of $v$ and $c$.","role":"user","type":"message"}],"instructions":null,"max_output_tokens":null,"max_tool_calls":null,"metadata":null,"model":null,"parallel_tool_calls":true,"previous_response_id":null,"prompt":null,"reasoning":null,"service_tier":null,"store":null,"temperature":null,"text":null,"tool_choice":"auto","tools":[],"top_logprobs":null,"top_p":null,"truncation":null,"user":null,"stream":null},"response":{"id":"resp_3hMgVzW0Lg2vQrLRy5TIMocVnih_t1qWsfsAyBmKkrZySD-PhEHi6TQv3U1omgt7_E5AbY21fGEtIYH2vFtov3_M6cKS-rCe4H_BNl635osOJe2ge5erLgcT9CinNagfH9tU8Kd6DVmo712zKHydfo61EtEVCu3FaNuvoAkBd2tNWPORsbm_d38eY1ynRdhKoSYeJJrYb_ZLEixHbLOB6f1MNVW1mqFMdBo-Te3s6rKnBe0WRBCWzb-ptOicPmHf1D7P8tlfRbIDy_ldmCaVC6-VlCZALyD3lNPmihDXbIrS0vON7vet-xu9n75UZ7tzzl7xIk4ZukcMAltegFq_yjmtmuXfWg==","created_at":1777570222.0,"error":null,"incomplete_details":null,"instructions":null,"metadata":{},"model":"gcp/google/gemini-2.5-pro","object":"response","output":[{"id":"rpHzaciQEICQ998Pme30gQE","content":[{"annotations":[],"text":"According to the theory of special relativity, time dilation is a difference in the elapsed time measured by two observers, either due to a velocity difference relative to each other, or by being at different locations relative to a gravitational field. In the context of relative velocity, this effect is described by the time dilation factor, commonly denoted by the Greek letter gamma ($\\gamma$).\n\nLet $\\Delta t_0$ be the proper time, which is the time interval measured in the reference frame of the moving object (the spaceship in this case). Let $\\Delta t$ be the time interval measured by an observer in a stationary reference frame (the Earth). The relationship between these two time intervals is given by the time dilation formula:\n$$ \\Delta t = \\gamma \\Delta t_0 $$\nThe time dilation factor $\\gamma$, also known as the Lorentz factor, depends on the relative speed $v$ between the two reference frames and the speed of light in a vacuum, $c$. The formula for $\\gamma$ is derived from the postulates of special relativity and is given by:\n$$ \\gamma = \\frac{1}{\\sqrt{1 - \\frac{v^2}{c^2}}} $$\nHere, $v$ is the speed of the spaceship relative to the Earth, and $c$ is the constant speed of light. Since $v < c$ for any object with mass, the term $\\frac{v^2}{c^2}$ is always less than 1, and thus $\\gamma$ is always a real number greater than or equal to 1. When $v=0$, $\\gamma=1$, and there is no time dilation. As $v$ approaches $c$, $\\gamma$ approaches infinity.\n\nThe question asks for the expression for the time dilation factor $\\gamma$ in terms of $v$ and $c$.\n\n$$ \\gamma = \\frac{1}{\\sqrt{1 - \\frac{v^2}{c^2}}} $$\n\n\\boxed{\\frac{1}{\\sqrt{1 - \\frac{v^2}{c^2}}}}","type":"output_text","logprobs":null}],"role":"assistant","status":"completed","type":"message"}],"parallel_tool_calls":false,"temperature":0.0,"tool_choice":"auto","tools":[],"top_p":null,"background":null,"conversation":null,"max_output_tokens":null,"max_tool_calls":null,"previous_response_id":null,"prompt":null,"prompt_cache_key":null,"reasoning":null,"safety_identifier":null,"service_tier":null,"status":"completed","text":{"format":null,"verbosity":null},"top_logprobs":null,"truncation":null,"usage":{"input_tokens":141,"input_tokens_details":{"cached_tokens":0,"audio_tokens":null,"text_tokens":141},"output_tokens":2262,"output_tokens_details":{"reasoning_tokens":1836,"text_tokens":426},"total_tokens":2403,"cost":null},"user":null,"store":null,"provider_specific_fields":{"traffic_type":"ON_DEMAND"}},"reward":1.0,"expected_answer":"\\boxed{1/\\sqrt{1-v^2/c^2}}","extracted_answer":"1/sqrt(1 - v**2/(c**2))","library_reward":1.0,"judge_evaluations":null,"_ng_task_index":4,"_ng_rollout_index":0,"agent_ref":{"type":"responses_api_agents","name":"physics_judge_simple_agent"}} -{"responses_create_params":{"background":null,"include":null,"input":[{"content":"Below is an open-ended problem in Physics. Please answer this problem adhering to the following rules:\n1. Please use LaTeX format to represent the variables and formulas used in the solution process and results.\n2. Please put the final answer(s) in \\boxed{}, note that the unit of the answer should not be included in \\boxed{}.\n3. If there are multiple final answers, please seperated them by commas in \\boxed{}, e.g., \\boxed{answer 1, answer 2}.\nProblem: A point charge $q$ is placed at the center of a hollow conducting sphere of inner radius $a$ and outer radius $b$. What is the surface charge density on the outer surface?","role":"user","type":"message"}],"instructions":null,"max_output_tokens":null,"max_tool_calls":null,"metadata":null,"model":null,"parallel_tool_calls":true,"previous_response_id":null,"prompt":null,"reasoning":null,"service_tier":null,"store":null,"temperature":null,"text":null,"tool_choice":"auto","tools":[],"top_logprobs":null,"top_p":null,"truncation":null,"user":null,"stream":null},"response":{"id":"resp_dkkwmbBa5Z0LOalY-7e7tYDyXpWHOz944gTYOVLX4Xu72nSXNGFPwaN-yu4V9yXLR55pKFmJdtFqqs4eKvLO73t7PhFeoAja4ymDBCqK4WOi7xiB4LmO7jJ1b1DRAzVWlnuJ2eCeDEhZJOsxmxuvn3PNgULRjHZbIZ3121XbWrEskHZ4RHjNkBTd4qmQ8cqYzjQeMqHmz4CvzthfELEOBjY9Wg5Xru5YGhmL_j8E_lEKAa7s7DtFgrhJwTheRmiEOL9RMKsYJHeX2OL0-7EjUGuG4pl33POAUaJrM-RAJIipj_5nOoaxPTYgDXYGKT0RpUNYuu6pW1IU9N3veSxgmdsJuA5Tkg==","created_at":1777570222.0,"error":null,"incomplete_details":null,"instructions":null,"metadata":{},"model":"gcp/google/gemini-2.5-pro","object":"response","output":[{"id":"rpHzadX4D9mN9LsPhOao2AY","content":[{"annotations":[],"text":"Here is a step-by-step solution to the problem.\n\nLet the hollow conducting sphere have an inner radius $a$ and an outer radius $b$. A point charge $q$ is placed at its center. We want to find the surface charge density on the outer surface, which we'll denote as $\\sigma_{outer}$.\n\n1. **Analyze the charge on the inner surface:**\n To determine the charge induced on the inner surface of the conductor, we can use Gauss's Law. Let's consider a spherical Gaussian surface with a radius $r$ such that it lies within the conducting material, i.e., $a < r < b$.\n In electrostatic equilibrium, the electric field $\\vec{E}$ inside a conductor is zero. Thus, for any point on our Gaussian surface, $\\vec{E} = 0$.\n\n Gauss's Law states:\n $\\oint \\vec{E} \\cdot d\\vec{A} = \\frac{Q_{enc}}{\\epsilon_0}$\n\n Since $\\vec{E} = 0$ everywhere on the Gaussian surface, the electric flux through the surface is zero:\n $\\oint \\vec{E} \\cdot d\\vec{A} = 0$\n\n This implies that the net charge enclosed by the Gaussian surface, $Q_{enc}$, must be zero. The enclosed charge is the sum of the point charge $q$ at the center and the charge induced on the inner surface of the sphere, let's call it $q_{inner}$.\n $Q_{enc} = q + q_{inner} = 0$\n\n Solving for $q_{inner}$, we find:\n $q_{inner} = -q$\n\n2. **Analyze the charge on the outer surface:**\n The problem does not state that the conducting sphere has a net charge, so we assume it is electrically neutral. The total charge on the conducting sphere, $Q_{sphere}$, is the sum of the charge on its inner surface ($q_{inner}$) and the charge on its outer surface ($q_{outer}$).\n For a neutral sphere, $Q_{sphere} = 0$.\n\n Therefore:\n $Q_{sphere} = q_{inner} + q_{outer} = 0$\n\n Substituting the value we found for $q_{inner}$:\n $-q + q_{outer} = 0$\n\n This gives us the charge on the outer surface:\n $q_{outer} = q$\n\n3. **Calculate the surface charge density on the outer surface:**\n The surface charge density, $\\sigma$, is defined as charge per unit area. The outer surface is a sphere with radius $b$. The area of the outer surface, $A_{outer}$, is:\n $A_{outer} = 4\\pi b^2$\n\n The surface charge density on the outer surface, $\\sigma_{outer}$, is the total charge on the outer surface divided by the area of the outer surface:\n $\\sigma_{outer} = \\frac{q_{outer}}{A_{outer}}$\n\n Substituting the values for $q_{outer}$ and $A_{outer}$:\n $\\sigma_{outer} = \\frac{q}{4\\pi b^2}$\n\nThe final answer for the surface charge density on the outer surface is $\\frac{q}{4\\pi b^2}$.\n\n\\boxed{\\frac{q}{4\\pi b^2}}","type":"output_text","logprobs":null}],"role":"assistant","status":"completed","type":"message"}],"parallel_tool_calls":false,"temperature":0.0,"tool_choice":"auto","tools":[],"top_p":null,"background":null,"conversation":null,"max_output_tokens":null,"max_tool_calls":null,"previous_response_id":null,"prompt":null,"prompt_cache_key":null,"reasoning":null,"safety_identifier":null,"service_tier":null,"status":"completed","text":{"format":null,"verbosity":null},"top_logprobs":null,"truncation":null,"usage":{"input_tokens":149,"input_tokens_details":{"cached_tokens":0,"audio_tokens":null,"text_tokens":149},"output_tokens":3073,"output_tokens_details":{"reasoning_tokens":2337,"text_tokens":736},"total_tokens":3222,"cost":null},"user":null,"store":null,"provider_specific_fields":{"traffic_type":"ON_DEMAND"}},"reward":1.0,"expected_answer":"\\boxed{q/(4\\pi b^2)}","extracted_answer":"q/((4*pi*b**2))","library_reward":1.0,"judge_evaluations":null,"_ng_task_index":2,"_ng_rollout_index":0,"agent_ref":{"type":"responses_api_agents","name":"physics_judge_simple_agent"}} -{"responses_create_params":{"background":null,"include":null,"input":[{"content":"Below is an open-ended problem in Physics. Please answer this problem adhering to the following rules:\n1. Please use LaTeX format to represent the variables and formulas used in the solution process and results.\n2. Please put the final answer(s) in \\boxed{}, note that the unit of the answer should not be included in \\boxed{}.\n3. If there are multiple final answers, please seperated them by commas in \\boxed{}, e.g., \\boxed{answer 1, answer 2}.\nProblem: An ideal monatomic gas at temperature $T$ undergoes an isothermal expansion from volume $V_1$ to $V_2$. What is the work done by the gas?","role":"user","type":"message"}],"instructions":null,"max_output_tokens":null,"max_tool_calls":null,"metadata":null,"model":null,"parallel_tool_calls":true,"previous_response_id":null,"prompt":null,"reasoning":null,"service_tier":null,"store":null,"temperature":null,"text":null,"tool_choice":"auto","tools":[],"top_logprobs":null,"top_p":null,"truncation":null,"user":null,"stream":null},"response":{"id":"resp_NMQ_5QYg8OD-VN1bQzKC-YVCQM7pQcPAAVuS0eB-cxrJnGjFO-ylF7Z92FBlYKaz0au611_7dR4c2WbU--XE-Vb4PV_MKOK81W04CphGDPnwfu-l7Cjtuj0tYXlBYuHE-93_mtYz-OihXkQ2zJmuNfYtMIJlxpEn6rM3OqowTsTsNuNTIVdm2tx2Y58-Gl60lKpSoModQTxnK02MOgVx2XpGZyDr7updxUgKxh6hCqtcRxHAabSh2rDuPeXxW9LK2KOuYKK9zS3awS0TBuatrYRXTgNeIO7CL6De2ppR4YdHTKPZWabrK_8xnbiotxfk4hFkVivU6xoh0uc9yUhg6NDAv3Vv_A==","created_at":1777570222.0,"error":null,"incomplete_details":null,"instructions":null,"metadata":{},"model":"gcp/google/gemini-2.5-pro","object":"response","output":[{"id":"rpHzaZH0Eeyhq8YP97_NyQg","content":[{"annotations":[],"text":"The work $W$ done by a gas on its surroundings during a change in volume is given by the integral of the pressure $P$ with respect to the volume $V$:\n$W = \\int_{V_1}^{V_2} P \\, dV$\n\nFor an ideal gas, the relationship between pressure, volume, and temperature is described by the ideal gas law:\n$PV = nRT$\nwhere $n$ is the number of moles of the gas, $R$ is the universal gas constant, and $T$ is the absolute temperature.\n\nWe can express the pressure $P$ in terms of the volume $V$ and other constants:\n$P = \\frac{nRT}{V}$\n\nThe problem states that the process is an isothermal expansion. This means the temperature $T$ of the gas remains constant throughout the process. We can substitute the expression for $P$ from the ideal gas law into the work integral. Since $n$, $R$, and $T$ are all constants during this process, we can move them outside the integral:\n$W = \\int_{V_1}^{V_2} \\frac{nRT}{V} \\, dV = nRT \\int_{V_1}^{V_2} \\frac{1}{V} \\, dV$\n\nNow, we evaluate the definite integral:\nThe integral of $\\frac{1}{V}$ with respect to $V$ is $\\ln|V|$.\n$W = nRT \\left[ \\ln|V| \\right]_{V_1}^{V_2}$\n\nSince volume is always a positive quantity, we can drop the absolute value signs:\n$W = nRT (\\ln(V_2) - \\ln(V_1))$\n\nUsing the properties of logarithms, specifically $\\ln(a) - \\ln(b) = \\ln(a/b)$, we can simplify the expression:\n$W = nRT \\ln\\left(\\frac{V_2}{V_1}\\right)$\n\nThis is the work done by the gas during an isothermal expansion from volume $V_1$ to $V_2$. Note that the information that the gas is monatomic is not required for this specific calculation, as the result holds for any ideal gas undergoing an isothermal process. For an expansion, $V_2 > V_1$, which means $\\frac{V_2}{V_1} > 1$, and therefore $\\ln\\left(\\frac{V_2}{V_1}\\right)$ is positive. This correctly indicates that the gas does positive work on its surroundings.\n\n\\boxed{nRT \\ln\\left(\\frac{V_2}{V_1}\\right)}","type":"output_text","logprobs":null}],"role":"assistant","status":"completed","type":"message"}],"parallel_tool_calls":false,"temperature":0.0,"tool_choice":"auto","tools":[],"top_p":null,"background":null,"conversation":null,"max_output_tokens":null,"max_tool_calls":null,"previous_response_id":null,"prompt":null,"prompt_cache_key":null,"reasoning":null,"safety_identifier":null,"service_tier":null,"status":"completed","text":{"format":null,"verbosity":null},"top_logprobs":null,"truncation":null,"usage":{"input_tokens":145,"input_tokens_details":{"cached_tokens":0,"audio_tokens":null,"text_tokens":145},"output_tokens":2928,"output_tokens_details":{"reasoning_tokens":2358,"text_tokens":570},"total_tokens":3073,"cost":null},"user":null,"store":null,"provider_specific_fields":{"traffic_type":"ON_DEMAND"}},"reward":1.0,"expected_answer":"\\boxed{nRT\\ln(V_2/V_1)}","extracted_answer":"n*r*t*log(v_2/v_1, E)","library_reward":1.0,"judge_evaluations":null,"_ng_task_index":1,"_ng_rollout_index":0,"agent_ref":{"type":"responses_api_agents","name":"physics_judge_simple_agent"}} +{"responses_create_params":{"background":null,"include":null,"input":[{"content":"Below is an open-ended problem in Physics. Please answer this problem adhering to the following rules:\n1. Please use LaTeX format to represent the variables and formulas used in the solution process and results.\n2. Please put the final answer(s) in \\boxed{}, note that the unit of the answer should not be included in \\boxed{}.\n3. If there are multiple final answers, please seperated them by commas in \\boxed{}, e.g., \\boxed{answer 1, answer 2}.\nProblem: What is the de Broglie wavelength of a particle with momentum $p$?","role":"user","type":"message"}],"instructions":null,"max_output_tokens":null,"max_tool_calls":null,"metadata":null,"model":null,"parallel_tool_calls":true,"previous_response_id":null,"prompt":null,"reasoning":null,"service_tier":null,"store":null,"temperature":null,"text":null,"tool_choice":"auto","tools":[],"top_logprobs":null,"top_p":null,"truncation":null,"user":null,"stream":null},"response":{"id":"resp_rWPwxggSyIMlON6Fp_EtQmrbpnV0OTYcVDJF0Sw6U6ygsjR4YRFRqgGWTbjpvL77rRISP6rrdeLMM9bFhgZw8t2LX24bT6otNpw03wYrL4VBRwjtoJhjCa6MC5WfidCNN3RV2AjSe8ICjD5V2JUNd3X4CXfvc_k21xA5Vvr3_DOYDgBhrE3bP9hn-m6OjxOLtgp1Punw7LXVC8T5t8vI29WW3q30LLtfKP5XcKucXJBgqwKL9kY7RO9VCqiod7yDfGxdcJIJl8_SgQOjYpS6GDLYrVQLi3D0VaQIQD2zrWu1fzdmnwpy2ok83zRoX--s7r62j2Ny38JbJmrmYzzJuMJnUmPzWg==","created_at":1777570222.0,"error":null,"incomplete_details":null,"instructions":null,"metadata":{},"model":"gcp/google/gemini-2.5-pro","object":"response","output":[{"id":"rpHzabyKD4-Pq8YPvqPZoAU","content":[{"annotations":[],"text":"According to the de Broglie hypothesis, all matter exhibits wave-like properties. The wavelength associated with a particle, known as the de Broglie wavelength, is related to its momentum.\n\nThe de Broglie wavelength, denoted by $\\lambda$, is given by the formula:\n$\\lambda = \\frac{h}{p}$\n\nwhere:\n- $\\lambda$ is the de Broglie wavelength.\n- $h$ is Planck's constant, a fundamental constant in physics with a value of approximately $h \\approx 6.626 \\times 10^{-34} \\text{ J} \\cdot \\text{s}$.\n- $p$ is the momentum of the particle.\n\nThus, the de Broglie wavelength of a particle with momentum $p$ is the ratio of Planck's constant to the momentum.\n\n\\boxed{\\frac{h}{p}}","type":"output_text","logprobs":null}],"role":"assistant","status":"completed","type":"message"}],"parallel_tool_calls":false,"temperature":0.0,"tool_choice":"auto","tools":[],"top_p":null,"background":null,"conversation":null,"max_output_tokens":null,"max_tool_calls":null,"previous_response_id":null,"prompt":null,"prompt_cache_key":null,"reasoning":null,"safety_identifier":null,"service_tier":null,"status":"completed","text":{"format":null,"verbosity":null},"top_logprobs":null,"truncation":null,"usage":{"input_tokens":125,"input_tokens_details":{"cached_tokens":0,"audio_tokens":null,"text_tokens":125},"output_tokens":1163,"output_tokens_details":{"reasoning_tokens":980,"text_tokens":183},"total_tokens":1288,"cost":null},"user":null,"store":null,"provider_specific_fields":{"traffic_type":"ON_DEMAND"}},"reward":1.0,"expected_answer":"\\boxed{h/p}","extracted_answer":"h/p","library_reward":1.0,"judge_evaluations":null,"domain":"quantum","_ng_task_index":3,"_ng_rollout_index":0,"agent_ref":{"type":"responses_api_agents","name":"physics_judge_simple_agent"}} +{"responses_create_params":{"background":null,"include":null,"input":[{"content":"Below is an open-ended problem in Physics. Please answer this problem adhering to the following rules:\n1. Please use LaTeX format to represent the variables and formulas used in the solution process and results.\n2. Please put the final answer(s) in \\boxed{}, note that the unit of the answer should not be included in \\boxed{}.\n3. If there are multiple final answers, please seperated them by commas in \\boxed{}, e.g., \\boxed{answer 1, answer 2}.\nProblem: A block of mass $m$ slides down a frictionless inclined plane of angle $\\theta$. What is its acceleration along the slope?","role":"user","type":"message"}],"instructions":null,"max_output_tokens":null,"max_tool_calls":null,"metadata":null,"model":null,"parallel_tool_calls":true,"previous_response_id":null,"prompt":null,"reasoning":null,"service_tier":null,"store":null,"temperature":null,"text":null,"tool_choice":"auto","tools":[],"top_logprobs":null,"top_p":null,"truncation":null,"user":null,"stream":null},"response":{"id":"resp_TEuGNod0Jgxvlowu1BMv-KlWV1pQEG7mZ0kde2eb8elfBYN9jV4yXNKMs8ZbS3GUPhkCu5Hyk0Ybkot1KY2j88IvvV_UpaaQ1IQWxRNpEQL7ImxaVK1sQ6Z9ui9lfCSXjhEE47wRFdZnT1GOtZEYp6UXv1Stb6Rda-SLxeApBetRoH_4c0O87CtqkftMUifTFPyikbGeealqQ8Ha6GHTStyT2GKDPqaUZx9ITWueKpsdxDiq8JucuLN9EOLIT4Rf-U1WRTe7R4UDu65pWne9xjA5oL0TRJP5J6Mv_Qr1_5r65YdrNiFH5ghmzdcEgvVmlzyjCAwqnQF-4wv7U7cc-3Om19PJKA==","created_at":1777570222.0,"error":null,"incomplete_details":null,"instructions":null,"metadata":{},"model":"gcp/google/gemini-2.5-pro","object":"response","output":[{"id":"rpHzabunEMuepucPsd6p6Ac","content":[{"annotations":[],"text":"Here is a step-by-step solution to the problem.\n\nLet the mass of the block be $m$ and the angle of the inclined plane be $\\theta$. The acceleration due to gravity is denoted by $g$.\n\n1. **Identify the forces acting on the block.**\n There are two forces acting on the block:\n * The gravitational force, $F_g = mg$, acting vertically downwards.\n * The normal force, $N$, exerted by the inclined plane on the block, acting perpendicular to the surface of the plane.\n Since the plane is frictionless, there is no friction force.\n\n2. **Choose a suitable coordinate system.**\n It is convenient to choose a tilted coordinate system where the x-axis is parallel to the inclined plane (pointing down the slope) and the y-axis is perpendicular to the inclined plane (pointing upwards).\n\n3. **Resolve the forces into components.**\n We resolve the gravitational force, $F_g$, into components parallel and perpendicular to the inclined plane.\n * The component of the gravitational force parallel to the slope is $F_{g,x} = mg \\sin(\\theta)$. This component pulls the block down the slope.\n * The component of the gravitational force perpendicular to the slope is $F_{g,y} = mg \\cos(\\theta)$. This component pushes the block against the plane.\n The normal force $N$ acts entirely along the positive y-axis.\n\n4. **Apply Newton's Second Law.**\n Newton's Second Law, $\\Sigma \\vec{F} = m\\vec{a}$, is applied along each axis of the coordinate system. Let $a$ be the acceleration of the block along the slope (in the x-direction).\n\n * **Along the y-axis (perpendicular to the slope):**\n The block does not accelerate in the direction perpendicular to the plane, so the net force in the y-direction is zero.\n $\\Sigma F_y = N - mg \\cos(\\theta) = 0$\n This gives $N = mg \\cos(\\theta)$.\n\n * **Along the x-axis (parallel to the slope):**\n The only force component acting along the x-axis is the parallel component of gravity.\n $\\Sigma F_x = mg \\sin(\\theta)$\n According to Newton's Second Law, the net force in the x-direction is also given by $\\Sigma F_x = ma$.\n Therefore, we can equate the two expressions:\n $ma = mg \\sin(\\theta)$\n\n5. **Solve for the acceleration.**\n To find the acceleration $a$, we can cancel the mass $m$ from both sides of the equation:\n $a = g \\sin(\\theta)$\n\nThis is the acceleration of the block sliding down the frictionless inclined plane.\n\nThe final answer is:\n\\boxed{g \\sin(\\theta)}","type":"output_text","logprobs":null}],"role":"assistant","status":"completed","type":"message"}],"parallel_tool_calls":false,"temperature":0.0,"tool_choice":"auto","tools":[],"top_p":null,"background":null,"conversation":null,"max_output_tokens":null,"max_tool_calls":null,"previous_response_id":null,"prompt":null,"prompt_cache_key":null,"reasoning":null,"safety_identifier":null,"service_tier":null,"status":"completed","text":{"format":null,"verbosity":null},"top_logprobs":null,"truncation":null,"usage":{"input_tokens":136,"input_tokens_details":{"cached_tokens":0,"audio_tokens":null,"text_tokens":136},"output_tokens":2149,"output_tokens_details":{"reasoning_tokens":1516,"text_tokens":633},"total_tokens":2285,"cost":null},"user":null,"store":null,"provider_specific_fields":{"traffic_type":"ON_DEMAND"}},"reward":1.0,"expected_answer":"\\boxed{g\\sin\\theta}","extracted_answer":"g*sin(theta)","library_reward":1.0,"judge_evaluations":null,"domain":"mechanics","_ng_task_index":0,"_ng_rollout_index":0,"agent_ref":{"type":"responses_api_agents","name":"physics_judge_simple_agent"}} +{"responses_create_params":{"background":null,"include":null,"input":[{"content":"Below is an open-ended problem in Physics. Please answer this problem adhering to the following rules:\n1. Please use LaTeX format to represent the variables and formulas used in the solution process and results.\n2. Please put the final answer(s) in \\boxed{}, note that the unit of the answer should not be included in \\boxed{}.\n3. If there are multiple final answers, please seperated them by commas in \\boxed{}, e.g., \\boxed{answer 1, answer 2}.\nProblem: A spaceship moves at speed $v$ relative to the Earth. Express the time dilation factor $\\gamma$ in terms of $v$ and $c$.","role":"user","type":"message"}],"instructions":null,"max_output_tokens":null,"max_tool_calls":null,"metadata":null,"model":null,"parallel_tool_calls":true,"previous_response_id":null,"prompt":null,"reasoning":null,"service_tier":null,"store":null,"temperature":null,"text":null,"tool_choice":"auto","tools":[],"top_logprobs":null,"top_p":null,"truncation":null,"user":null,"stream":null},"response":{"id":"resp_3hMgVzW0Lg2vQrLRy5TIMocVnih_t1qWsfsAyBmKkrZySD-PhEHi6TQv3U1omgt7_E5AbY21fGEtIYH2vFtov3_M6cKS-rCe4H_BNl635osOJe2ge5erLgcT9CinNagfH9tU8Kd6DVmo712zKHydfo61EtEVCu3FaNuvoAkBd2tNWPORsbm_d38eY1ynRdhKoSYeJJrYb_ZLEixHbLOB6f1MNVW1mqFMdBo-Te3s6rKnBe0WRBCWzb-ptOicPmHf1D7P8tlfRbIDy_ldmCaVC6-VlCZALyD3lNPmihDXbIrS0vON7vet-xu9n75UZ7tzzl7xIk4ZukcMAltegFq_yjmtmuXfWg==","created_at":1777570222.0,"error":null,"incomplete_details":null,"instructions":null,"metadata":{},"model":"gcp/google/gemini-2.5-pro","object":"response","output":[{"id":"rpHzaciQEICQ998Pme30gQE","content":[{"annotations":[],"text":"According to the theory of special relativity, time dilation is a difference in the elapsed time measured by two observers, either due to a velocity difference relative to each other, or by being at different locations relative to a gravitational field. In the context of relative velocity, this effect is described by the time dilation factor, commonly denoted by the Greek letter gamma ($\\gamma$).\n\nLet $\\Delta t_0$ be the proper time, which is the time interval measured in the reference frame of the moving object (the spaceship in this case). Let $\\Delta t$ be the time interval measured by an observer in a stationary reference frame (the Earth). The relationship between these two time intervals is given by the time dilation formula:\n$$ \\Delta t = \\gamma \\Delta t_0 $$\nThe time dilation factor $\\gamma$, also known as the Lorentz factor, depends on the relative speed $v$ between the two reference frames and the speed of light in a vacuum, $c$. The formula for $\\gamma$ is derived from the postulates of special relativity and is given by:\n$$ \\gamma = \\frac{1}{\\sqrt{1 - \\frac{v^2}{c^2}}} $$\nHere, $v$ is the speed of the spaceship relative to the Earth, and $c$ is the constant speed of light. Since $v < c$ for any object with mass, the term $\\frac{v^2}{c^2}$ is always less than 1, and thus $\\gamma$ is always a real number greater than or equal to 1. When $v=0$, $\\gamma=1$, and there is no time dilation. As $v$ approaches $c$, $\\gamma$ approaches infinity.\n\nThe question asks for the expression for the time dilation factor $\\gamma$ in terms of $v$ and $c$.\n\n$$ \\gamma = \\frac{1}{\\sqrt{1 - \\frac{v^2}{c^2}}} $$\n\n\\boxed{\\frac{1}{\\sqrt{1 - \\frac{v^2}{c^2}}}}","type":"output_text","logprobs":null}],"role":"assistant","status":"completed","type":"message"}],"parallel_tool_calls":false,"temperature":0.0,"tool_choice":"auto","tools":[],"top_p":null,"background":null,"conversation":null,"max_output_tokens":null,"max_tool_calls":null,"previous_response_id":null,"prompt":null,"prompt_cache_key":null,"reasoning":null,"safety_identifier":null,"service_tier":null,"status":"completed","text":{"format":null,"verbosity":null},"top_logprobs":null,"truncation":null,"usage":{"input_tokens":141,"input_tokens_details":{"cached_tokens":0,"audio_tokens":null,"text_tokens":141},"output_tokens":2262,"output_tokens_details":{"reasoning_tokens":1836,"text_tokens":426},"total_tokens":2403,"cost":null},"user":null,"store":null,"provider_specific_fields":{"traffic_type":"ON_DEMAND"}},"reward":1.0,"expected_answer":"\\boxed{1/\\sqrt{1-v^2/c^2}}","extracted_answer":"1/sqrt(1 - v**2/(c**2))","library_reward":1.0,"judge_evaluations":null,"domain":"relativity","_ng_task_index":4,"_ng_rollout_index":0,"agent_ref":{"type":"responses_api_agents","name":"physics_judge_simple_agent"}} +{"responses_create_params":{"background":null,"include":null,"input":[{"content":"Below is an open-ended problem in Physics. Please answer this problem adhering to the following rules:\n1. Please use LaTeX format to represent the variables and formulas used in the solution process and results.\n2. Please put the final answer(s) in \\boxed{}, note that the unit of the answer should not be included in \\boxed{}.\n3. If there are multiple final answers, please seperated them by commas in \\boxed{}, e.g., \\boxed{answer 1, answer 2}.\nProblem: A point charge $q$ is placed at the center of a hollow conducting sphere of inner radius $a$ and outer radius $b$. What is the surface charge density on the outer surface?","role":"user","type":"message"}],"instructions":null,"max_output_tokens":null,"max_tool_calls":null,"metadata":null,"model":null,"parallel_tool_calls":true,"previous_response_id":null,"prompt":null,"reasoning":null,"service_tier":null,"store":null,"temperature":null,"text":null,"tool_choice":"auto","tools":[],"top_logprobs":null,"top_p":null,"truncation":null,"user":null,"stream":null},"response":{"id":"resp_dkkwmbBa5Z0LOalY-7e7tYDyXpWHOz944gTYOVLX4Xu72nSXNGFPwaN-yu4V9yXLR55pKFmJdtFqqs4eKvLO73t7PhFeoAja4ymDBCqK4WOi7xiB4LmO7jJ1b1DRAzVWlnuJ2eCeDEhZJOsxmxuvn3PNgULRjHZbIZ3121XbWrEskHZ4RHjNkBTd4qmQ8cqYzjQeMqHmz4CvzthfELEOBjY9Wg5Xru5YGhmL_j8E_lEKAa7s7DtFgrhJwTheRmiEOL9RMKsYJHeX2OL0-7EjUGuG4pl33POAUaJrM-RAJIipj_5nOoaxPTYgDXYGKT0RpUNYuu6pW1IU9N3veSxgmdsJuA5Tkg==","created_at":1777570222.0,"error":null,"incomplete_details":null,"instructions":null,"metadata":{},"model":"gcp/google/gemini-2.5-pro","object":"response","output":[{"id":"rpHzadX4D9mN9LsPhOao2AY","content":[{"annotations":[],"text":"Here is a step-by-step solution to the problem.\n\nLet the hollow conducting sphere have an inner radius $a$ and an outer radius $b$. A point charge $q$ is placed at its center. We want to find the surface charge density on the outer surface, which we'll denote as $\\sigma_{outer}$.\n\n1. **Analyze the charge on the inner surface:**\n To determine the charge induced on the inner surface of the conductor, we can use Gauss's Law. Let's consider a spherical Gaussian surface with a radius $r$ such that it lies within the conducting material, i.e., $a < r < b$.\n In electrostatic equilibrium, the electric field $\\vec{E}$ inside a conductor is zero. Thus, for any point on our Gaussian surface, $\\vec{E} = 0$.\n\n Gauss's Law states:\n $\\oint \\vec{E} \\cdot d\\vec{A} = \\frac{Q_{enc}}{\\epsilon_0}$\n\n Since $\\vec{E} = 0$ everywhere on the Gaussian surface, the electric flux through the surface is zero:\n $\\oint \\vec{E} \\cdot d\\vec{A} = 0$\n\n This implies that the net charge enclosed by the Gaussian surface, $Q_{enc}$, must be zero. The enclosed charge is the sum of the point charge $q$ at the center and the charge induced on the inner surface of the sphere, let's call it $q_{inner}$.\n $Q_{enc} = q + q_{inner} = 0$\n\n Solving for $q_{inner}$, we find:\n $q_{inner} = -q$\n\n2. **Analyze the charge on the outer surface:**\n The problem does not state that the conducting sphere has a net charge, so we assume it is electrically neutral. The total charge on the conducting sphere, $Q_{sphere}$, is the sum of the charge on its inner surface ($q_{inner}$) and the charge on its outer surface ($q_{outer}$).\n For a neutral sphere, $Q_{sphere} = 0$.\n\n Therefore:\n $Q_{sphere} = q_{inner} + q_{outer} = 0$\n\n Substituting the value we found for $q_{inner}$:\n $-q + q_{outer} = 0$\n\n This gives us the charge on the outer surface:\n $q_{outer} = q$\n\n3. **Calculate the surface charge density on the outer surface:**\n The surface charge density, $\\sigma$, is defined as charge per unit area. The outer surface is a sphere with radius $b$. The area of the outer surface, $A_{outer}$, is:\n $A_{outer} = 4\\pi b^2$\n\n The surface charge density on the outer surface, $\\sigma_{outer}$, is the total charge on the outer surface divided by the area of the outer surface:\n $\\sigma_{outer} = \\frac{q_{outer}}{A_{outer}}$\n\n Substituting the values for $q_{outer}$ and $A_{outer}$:\n $\\sigma_{outer} = \\frac{q}{4\\pi b^2}$\n\nThe final answer for the surface charge density on the outer surface is $\\frac{q}{4\\pi b^2}$.\n\n\\boxed{\\frac{q}{4\\pi b^2}}","type":"output_text","logprobs":null}],"role":"assistant","status":"completed","type":"message"}],"parallel_tool_calls":false,"temperature":0.0,"tool_choice":"auto","tools":[],"top_p":null,"background":null,"conversation":null,"max_output_tokens":null,"max_tool_calls":null,"previous_response_id":null,"prompt":null,"prompt_cache_key":null,"reasoning":null,"safety_identifier":null,"service_tier":null,"status":"completed","text":{"format":null,"verbosity":null},"top_logprobs":null,"truncation":null,"usage":{"input_tokens":149,"input_tokens_details":{"cached_tokens":0,"audio_tokens":null,"text_tokens":149},"output_tokens":3073,"output_tokens_details":{"reasoning_tokens":2337,"text_tokens":736},"total_tokens":3222,"cost":null},"user":null,"store":null,"provider_specific_fields":{"traffic_type":"ON_DEMAND"}},"reward":1.0,"expected_answer":"\\boxed{q/(4\\pi b^2)}","extracted_answer":"q/((4*pi*b**2))","library_reward":1.0,"judge_evaluations":null,"domain":"electromagnetism","_ng_task_index":2,"_ng_rollout_index":0,"agent_ref":{"type":"responses_api_agents","name":"physics_judge_simple_agent"}} +{"responses_create_params":{"background":null,"include":null,"input":[{"content":"Below is an open-ended problem in Physics. Please answer this problem adhering to the following rules:\n1. Please use LaTeX format to represent the variables and formulas used in the solution process and results.\n2. Please put the final answer(s) in \\boxed{}, note that the unit of the answer should not be included in \\boxed{}.\n3. If there are multiple final answers, please seperated them by commas in \\boxed{}, e.g., \\boxed{answer 1, answer 2}.\nProblem: An ideal monatomic gas at temperature $T$ undergoes an isothermal expansion from volume $V_1$ to $V_2$. What is the work done by the gas?","role":"user","type":"message"}],"instructions":null,"max_output_tokens":null,"max_tool_calls":null,"metadata":null,"model":null,"parallel_tool_calls":true,"previous_response_id":null,"prompt":null,"reasoning":null,"service_tier":null,"store":null,"temperature":null,"text":null,"tool_choice":"auto","tools":[],"top_logprobs":null,"top_p":null,"truncation":null,"user":null,"stream":null},"response":{"id":"resp_NMQ_5QYg8OD-VN1bQzKC-YVCQM7pQcPAAVuS0eB-cxrJnGjFO-ylF7Z92FBlYKaz0au611_7dR4c2WbU--XE-Vb4PV_MKOK81W04CphGDPnwfu-l7Cjtuj0tYXlBYuHE-93_mtYz-OihXkQ2zJmuNfYtMIJlxpEn6rM3OqowTsTsNuNTIVdm2tx2Y58-Gl60lKpSoModQTxnK02MOgVx2XpGZyDr7updxUgKxh6hCqtcRxHAabSh2rDuPeXxW9LK2KOuYKK9zS3awS0TBuatrYRXTgNeIO7CL6De2ppR4YdHTKPZWabrK_8xnbiotxfk4hFkVivU6xoh0uc9yUhg6NDAv3Vv_A==","created_at":1777570222.0,"error":null,"incomplete_details":null,"instructions":null,"metadata":{},"model":"gcp/google/gemini-2.5-pro","object":"response","output":[{"id":"rpHzaZH0Eeyhq8YP97_NyQg","content":[{"annotations":[],"text":"The work $W$ done by a gas on its surroundings during a change in volume is given by the integral of the pressure $P$ with respect to the volume $V$:\n$W = \\int_{V_1}^{V_2} P \\, dV$\n\nFor an ideal gas, the relationship between pressure, volume, and temperature is described by the ideal gas law:\n$PV = nRT$\nwhere $n$ is the number of moles of the gas, $R$ is the universal gas constant, and $T$ is the absolute temperature.\n\nWe can express the pressure $P$ in terms of the volume $V$ and other constants:\n$P = \\frac{nRT}{V}$\n\nThe problem states that the process is an isothermal expansion. This means the temperature $T$ of the gas remains constant throughout the process. We can substitute the expression for $P$ from the ideal gas law into the work integral. Since $n$, $R$, and $T$ are all constants during this process, we can move them outside the integral:\n$W = \\int_{V_1}^{V_2} \\frac{nRT}{V} \\, dV = nRT \\int_{V_1}^{V_2} \\frac{1}{V} \\, dV$\n\nNow, we evaluate the definite integral:\nThe integral of $\\frac{1}{V}$ with respect to $V$ is $\\ln|V|$.\n$W = nRT \\left[ \\ln|V| \\right]_{V_1}^{V_2}$\n\nSince volume is always a positive quantity, we can drop the absolute value signs:\n$W = nRT (\\ln(V_2) - \\ln(V_1))$\n\nUsing the properties of logarithms, specifically $\\ln(a) - \\ln(b) = \\ln(a/b)$, we can simplify the expression:\n$W = nRT \\ln\\left(\\frac{V_2}{V_1}\\right)$\n\nThis is the work done by the gas during an isothermal expansion from volume $V_1$ to $V_2$. Note that the information that the gas is monatomic is not required for this specific calculation, as the result holds for any ideal gas undergoing an isothermal process. For an expansion, $V_2 > V_1$, which means $\\frac{V_2}{V_1} > 1$, and therefore $\\ln\\left(\\frac{V_2}{V_1}\\right)$ is positive. This correctly indicates that the gas does positive work on its surroundings.\n\n\\boxed{nRT \\ln\\left(\\frac{V_2}{V_1}\\right)}","type":"output_text","logprobs":null}],"role":"assistant","status":"completed","type":"message"}],"parallel_tool_calls":false,"temperature":0.0,"tool_choice":"auto","tools":[],"top_p":null,"background":null,"conversation":null,"max_output_tokens":null,"max_tool_calls":null,"previous_response_id":null,"prompt":null,"prompt_cache_key":null,"reasoning":null,"safety_identifier":null,"service_tier":null,"status":"completed","text":{"format":null,"verbosity":null},"top_logprobs":null,"truncation":null,"usage":{"input_tokens":145,"input_tokens_details":{"cached_tokens":0,"audio_tokens":null,"text_tokens":145},"output_tokens":2928,"output_tokens_details":{"reasoning_tokens":2358,"text_tokens":570},"total_tokens":3073,"cost":null},"user":null,"store":null,"provider_specific_fields":{"traffic_type":"ON_DEMAND"}},"reward":1.0,"expected_answer":"\\boxed{nRT\\ln(V_2/V_1)}","extracted_answer":"n*r*t*log(v_2/v_1, E)","library_reward":1.0,"judge_evaluations":null,"domain":"thermodynamics","_ng_task_index":1,"_ng_rollout_index":0,"agent_ref":{"type":"responses_api_agents","name":"physics_judge_simple_agent"}} diff --git a/resources_servers/physics_judge/task_data.py b/resources_servers/physics_judge/task_data.py index a7516799862..59e5a81c0dd 100644 --- a/resources_servers/physics_judge/task_data.py +++ b/resources_servers/physics_judge/task_data.py @@ -2,12 +2,11 @@ # SPDX-License-Identifier: Apache-2.0 """Task-data schema for the physics_judge server. -Heir of math_with_judge: physics_judge defines no request model of its own — verify() and the -wire contract are the inherited ``LibraryJudgeMathVerifyRequest`` (question + expected_answer, -both required). Committed rows add physics metadata on top: ``domain`` is read defensively at -metrics-aggregation time (compute_metrics, app.py:240-255) for per-domain subsets, while -``difficulty``/``answer_type``/``language``/``solution`` are pass-through provenance written by -benchmarks/physics/prepare.py and never read by server code. +Heir of math_with_judge: the wire contract is ``PhysicsJudgeVerifyRequest``, the inherited +question + expected_answer (both required) plus an optional ``domain``, which verify() echoes so +compute_metrics can build per-domain subsets from the verify rows. ``difficulty``/``answer_type``/ +``language``/``solution`` are pass-through provenance written by benchmarks/physics/prepare.py and +never read by server code. """ from typing import Optional diff --git a/resources_servers/physics_judge/tests/test_app.py b/resources_servers/physics_judge/tests/test_app.py index 637bee4bd98..29f8cf93fbb 100644 --- a/resources_servers/physics_judge/tests/test_app.py +++ b/resources_servers/physics_judge/tests/test_app.py @@ -21,10 +21,12 @@ """ import json +from pathlib import Path from typing import Any from unittest.mock import AsyncMock, MagicMock from pytest import approx, fixture, raises +from starlette.testclient import TestClient from nemo_gym.config_types import ModelServerRef from nemo_gym.judge import JudgeError @@ -361,6 +363,62 @@ def test_no_domain_field_skips_subset_metrics(self, server: PhysicsJudgeResource assert not any("/" in k and not k.startswith("pass@") and not k.startswith("majority@") for k in metrics) +class TestDomainThroughVerify: + """`domain` must survive the real /verify route, since compute_metrics reads it from verify rows.""" + + @fixture + def client(self, config: PhysicsJudgeResourcesServerConfig) -> TestClient: + # Symbolic-only, so the gold answer scores without a judge call. + server = PhysicsJudgeResourcesServer( + config=config.model_copy(update={"should_use_judge": False}), + server_client=MagicMock(spec=ServerClient), + ) + with TestClient(server.setup_webserver()) as client: + yield client + + def _verify_all(self, client: TestClient, rows: list[dict[str, Any]]) -> list[dict[str, Any]]: + verified = [] + for task_index, row in enumerate(rows): + body = {**row, "response": _make_text_response(row["expected_answer"])} + response = client.post("/verify", json=body) + assert response.status_code == 200, response.text + verified.append({**response.json(), "_ng_task_index": task_index, "_ng_rollout_index": 0}) + return verified + + def test_shipped_rows_keep_domain_and_get_per_domain_metrics(self, client: TestClient) -> None: + example = Path(__file__).parents[1] / "data" / "example.jsonl" + rows = [json.loads(line) for line in example.read_text().splitlines()] + domains = [row["domain"] for row in rows] + assert len(set(domains)) == 5 + + verified = self._verify_all(client, rows) + assert [row.get("domain") for row in verified] == domains + + response = client.post("/aggregate_metrics", json={"verify_responses": verified}) + assert response.status_code == 200, response.text + metrics = response.json()["agent_metrics"] + assert metrics["mean/reward"] == approx(1.0) + for domain in domains: + assert metrics[f"{domain}/pass@1/symbolic_accuracy"] == approx(100.0) + assert metrics[f"{domain}/pass@1/no_answer"] == approx(0.0) + + def test_row_without_domain_gets_no_subset_metrics(self, client: TestClient) -> None: + row = { + "responses_create_params": {"input": [{"role": "user", "content": "What is 2+2?"}]}, + "question": "What is 2+2?", + "expected_answer": r"\boxed{4}", + } + verified = self._verify_all(client, [row]) + assert verified[0].get("domain") is None + + # A null `domain` must add nothing over a row that has no `domain` key at all. + without_key = [{k: v for k, v in verified[0].items() if k != "domain"}] + metrics = client.post("/aggregate_metrics", json={"verify_responses": verified}).json()["agent_metrics"] + baseline = client.post("/aggregate_metrics", json={"verify_responses": without_key}).json()["agent_metrics"] + assert metrics["mean/reward"] == approx(1.0) + assert metrics.keys() == baseline.keys() + + class TestGetKeyMetrics: def test_includes_token_means_when_present(self, server: PhysicsJudgeResourcesServer) -> None: agent_metrics = {