I have read in Zeemansky's physics   $dQ=dU+pdV$ for first law of thermodynamics  But when I came across another book of thermal physics,it says  $δQ= dU +pdV$. So what us the difference ?
10960
I have read the thread regarding , but it does not answer my question.  I am confused about the notation for change in Physics. In Mathematics, $\delta$ and $\Delta$ essentially refer to the same thing, i.e., change. This means that $\Delta x = x_1 - x_2 = \delta x$. The difference between $\delta$ and $d$ is also clear and distinct in differential calculus. We know that $\frac{dy}{dx}$ is always an operator and not a fraction, whereas $\frac{\delta y}{\delta x}$ is an infinitesimal change.   In Physics, however, the distinction is not as clear. Can anyone offer a clearer picture?
62584
I would like to know the physical meaning of the , if there is any? I've used it in thermodynamics and classical mechanics and it seemed only a change of coordinates?
43513
,I got one confusion when reading Goldstein's Classical Mechanics (page 20, third edition). After getting the equation $$ \sum \left\{\left[\frac{\mathrm{d}}{\mathrm{d}t}{\left(\frac{\partial T}{\partial \dot{q_{j}}} \right)} - \frac{\partial T}{\partial q_{j} } \right] - Q_{j} \right\} \ \delta q_{j} = 0 $$ then it says that     Note that in a system of Cartesian coordinates the partial derivative of $T$ with respect to $q_{j}$ vanishes. Thus, speaking in the language of differential geometry, this term arises from the curvature of the coordinates $q_{j}$. In polar coordinates, e.g., it is in the partial derivative of $T$ with respect to an angle coordinate that the centripetal acceleration term appears.   My question is:  Is the above statement general, i.e., that the kinetic energy $T$ does not depend on the position. I wonder why velocity can't depend on the particle's position vector. I mean, why couldn't we have cases where $\vec{v} = \vec{v}(\vec r ,t)$, so that the kinetic energy depends on $q_{j}$ or $\vec{r}$?    I will use the example of the second answer in this post   the second answer,   suppose we have a velocity distribution of gas whose function is $\vec {v}=\vec {v}(x,y,z,t)$  then the partial derivative with respect to $x$ is  $$\frac{\partial T} {\partial x}=mv\frac{\partial v} {\partial x}$$  since $v$ depends on $x$, I think in this case. $\frac{\partial T}{\partial q_{j} }$ does not vanish.
70630
Basically, which is the correct formula for thermal energy, and is this the same as kinetic energy? My notes are pretty conflicting on this topic, and I'm getting pretty confused.
108697
I am very new to relativity, and as I was going through Resnick &amp; Halliday, I saw that momentum was no longer $mv$, rather $\gamma mv$. This was the proof:     $$p =  mv = m \frac{\delta x}{\delta t_0}  $$   where $\delta t_0$ is the proper infinitesimal time interval to cover $\delta x$ distance.     Using the time dilation formula, $\delta t = \gamma \delta t_0$ we can then write    $$ p = m \frac{\delta x}{\delta t_0} = m \frac{\delta x}{\delta t} \frac{\delta t}{\delta t_0} = \gamma m \frac{\delta x}{\delta t} $$   However, since $ \frac{\delta x}{\delta t} $ is just the [observed] particle velocity,    $$ p = \gamma mv $$   On the other hand, when I went on to Feynmann's book I saw that there he talked about using "the new m" (the relativistic mass) to derive all the equations and said      $$p = mv = \frac{m_0 v}{(1 - \frac{v^2}{c^2})} = \gamma m_0 v $$   My problem is that if both these proofs are true, then why doesn't one consider the other's argument, id est, why aren't both of these factors considered together ? If the mass is really changing, why doesn't Resnick &amp; Halliday account for it? That way the momentum would turn out to be $p = \gamma^2 mv$  Am I missing something here?
30159
What is meant by ? My professor tells me "heat content". That literally makes no sense. Heat content, to me, means internal energy. But clearly, that is not what enthalpy is, considering: $H=U+PV$ (and either way, they would not have had two words mean the same thing). Then, I understand that $ΔH=Q_{p}$. This statement is a mathematical formulation of the statement: "At constant pressure, enthalpy change may be interpreted as heat." Other than this, I have no idea, what $H$ or $ΔH$ means.  So what does $H$ mean?
25325
The  in fluid mechanics states that  $$ \frac{\partial\rho}{\partial t} + \nabla\cdot(ρ\mathbf u)=0 $$ Can you explain to me what is the physical meaning of each term of the equation and what are the direct outcomes?  Keep in mind that i do not have much knowledge about differential calculus. Thank you! EDIT: After receiving some answers(for those who want to find out the answer,check the comments),i have only one question left.If the density gets bigger and bigger as time passes,then the first term of the equation is positive.What happens with the second term?What changes in that control volume?I mean,the density in the CV gets bigger,so what happens to velocity(intuitively)?I think that the velocity must get bigger in order to the fluid to get out of the CV.But that means that the second term is also positive and the equation is not  right.So what is wrong with my thinking?
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Please refer to my school textbook pg48 (of the book, and not the pdf counter) here:   My doubt is in this context: (right side column)  $a = dv/dt = v dv/dx$  then integrating $v$ with respect to $dv$, and $a$ with respect to $dx$.  Now, when we integrate $a$, either we say it is a constant, and give $a\times(x-x_o)$ as a result of the integration, or we say it is not constant, and is a function of time, in which case, it can not be integrated like this. But the book integrates it like this, and then it is written:     The advantage of this method is that it can be used for motion with non-uniform acceleration also.   Can you please explain how it can be used with non uniform acceleration while it has been assumed while integrating that $a$ is constant? And moreover, does this line seem to apply to only this derivation, or does it apply to other two before it as well?
23529
What is the exact use of the symbols $\partial$, $\delta$ and $\mathrm{d}$ in derivatives in physics? How are they different and when are they used? It would be nice to get that settled once and for all.  $$\frac{\partial y}{\partial x}, \frac{\delta y}{\delta x}, \frac{\mathrm{d} y}{\mathrm{d} x}$$   For what I know, $\mathrm{d}$ is used as a small infinitisemal change (and I guess the straight-up letter $\mathrm{d}$ is usual notation instead of italic $d$, simply to tell the difference from a variable). Of course we also have the big delta $\Delta$ to describe a finite (non-negligible) difference. And I have some vague idea that $\partial$ is used for partial derivatives in case of e.g. three-dimensional variables. Same goes for $\delta$, which I would have sworn was the same as $\partial$ until reading this answer on Math.SE:    Then to make the confusion total I noticed an equation like $\delta Q=\mathrm{d}U+\delta W$ and read in a physics text book that:     The fact that the amount of heat [added between two states] is dependent on the path is indicated by the symbol $\delta$...   So it seems $\delta$ means something more? The text book continues and says that:     a function [like the change in internal energy] is called a state function and its change is indicated by the symbol $\mathrm{d}$...   Here I am unsure of exactly why a $\mathrm d$ refers to a state function.   So to sum it up: down to the bone of it, what is $\delta$, $\partial$ and $\mathrm{d}$ exactly, when we are talking derivatives in physics.  Addition  Especially when reading a mathematical process on a physical equation like this procedure:  $$\delta Q=\mathrm{d}U+p\mathrm{d}V \Rightarrow\\ Q=\Delta U+\int_1^2 p \mathrm{d}V$$  It appears that $\delta$ and $\mathrm{d}$ are the same thing. An integral operation handles it the same way apparently?
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If we take heat capacity to be defined as "the ratio of the heat added to the temperature rise":  $$ C=\frac{\text{d} Q_{rev}}{\text{d}\theta}$$  then this leads me to ask:  can this ever be negative? That is to say, are there any materials which cool as you add heat to them?
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In thermodynamics, the first law can be written in differential form as $$dU = \delta Q - \delta W$$ Here, $dU$ is the differential $1$-form of the internal energy but $\delta Q$ and $\delta W$ are inexact differentials, which is emphasized with the replacement of $d$ with $\delta $.   My question is why we regard heat (or work) as differential forms? Suppose our system can be described with state variables $(x_1,\ \cdots ,\ x_n)$. To my understanding, a general $1$-form is written as $$df = f_1\ dx_1 + f_2\ dx_2 + \cdots + f_n\ dx_n$$ In particular, differential forms are linear functionals of our state variables. Is there any good reason to presuppose that $\delta Q$ is linear in the state variables?   In other words, if the infinitesimal heat transferred when moving from state $\mathbf{x}$ to $\mathbf{x} + d\mathbf{x}_1$ is $\delta Q_1$ and from $\mathbf{x}$ to $\mathbf{x} + d\mathbf{x}_2$ is $\delta Q_2$, is there any physical reason why the heat transferred from $\mathbf{x}$ to $\mathbf{x} + d\mathbf{x}_1 + d\mathbf{x}_2$ should be $\delta Q_1 + \delta Q_2$?  I apologize if the question is a bit unclear, I am still trying to formulate some of these ideas in my head.  P.S. I know that for quasi-static processes, we have $\delta Q = T\ dS$ (and $\delta W = p\ dV$) so have shown $\delta Q$ is a differential form in this case. I guess my question is about non-quasi-static processes in general.
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What is the potential diff between two plates of charges $q_1$ and $q_2$ ($q_1$ $&gt;$ $q_2$)? I know the equation is $V = \frac{q_1-q_2}{2C}$ where $C$ is the capacitance of the system but how has this equation been derived?   I apologize if this is a duplicate question. Google searches with different search terms didn't yield any answers.
53337
I was trying to derive equation (65) in the following review:    This slightly unusual then usual classical mechanics because it includes a variation of time also, $\delta t$.  Usually one would define, $$\delta S~=~ \int \left[ L \left(\tilde{q}(t),\dot{\tilde{q}}(t),t\right) -  L \left(q(t),\dot{q}(t),t\right)\right] dt , $$ where,  $$\tilde{q}~=~q+\delta q. $$  We have then (before applying int by parts),  $$ \delta S ~=~ \int \left(\frac{\partial L}{\partial q}\delta q +\frac{\partial L}{\partial \dot q}\delta \dot q\right)dt.$$  How does one proceed if both $q$ and $t$ vary and further that $q$ depends on a varying $t$?  Is the definition now  $$\delta S~=~ \int \left[ L \left(\tilde{q}(t+\delta t),\dot{\tilde{q}}( t+\delta t), t+\delta t\right) -  L \left(q(t),\dot{q}(t),t\right)\right] dt~?  $$  If so how does one proceed?
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why: $$\int_0^1\int_{u(t)}^{u(t)+w(t)} f(t,v(t)) dv dt = \int_0^1 f(t,u(t)+\theta w(t))w(t) dt; ~~\theta\in[0,1] $$  how to get this ?  Please help me   Thank you.
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Entropy is defined in my book as $\Delta\ S = \frac{Q}{T}$. To derive the formula it says that entropy should be directly proportional to the heat energy as with more energy the particles would be flying all over the place more rapidly. This makes sense. But then it says that entropy must be inversely proportional to the temperature at which it is added. To be precise  Heat added to a system at a lower temperature causes higher entropy increase than heat added to the same system at a higher temperature.  How does this make intuitive sense? EDIT 1: I found an answer . I think this makes sense. Can anyone read it and tell me if it's correct or not?  Rough intuitive answer: Adding heat to a system increases the system's total energy. This gives more kinetic energy to distribute among the particles in the system, increasing the size of the system's phase space and hence its entropy. However, since momentum is proportional to the square root of kinetic energy, it is clear that the total momentum phase space cannot grow exponentially as you add more kinetic energy; so the rate of growth of its logarithm must taper off. Since entropy is proportional to the log of the size of the phase space, we see that each additional unit of heat must increase the entropy by less and less. Now, for technical details. The reason why the relationship is precisely $dS = \delta q/T$ rather than some more complicated relationship that depends on the nature of the system is that temperature is defined such that this is true.  We know that during first-order phase transitions, the temperature actually stays constant as you add heat! This corresponds to a region where adding heat actually does cause an exponential increase in phase space. Finally, note that some systems actually have negative temperature, and adding more heat to them actually makes the temperature even more negative; so adding heat actually decreases the phase space. So the intuitive answer in the previous paragraph shouldn't be taken too seriously.  EDIT 2: Moreover,  answer explains why there needs to be an additional factor of temperature along with heat in the definition of entropy so as to be consistent with the second law.  This is why heat flows from hot objects to cold objects: the entropy change of the hot object is negative, while that of the cold object is positive. The magnitude of entropy change for the cold object is greater. $\Delta S = \frac{Q}{T_c}-\frac{Q}{T_h} &gt; 0$ as $T_h &gt; T_c$  Keep in mind that entropy increases with temperature. This can be understood intuitively in the classical picture, as you mention. However, at higher temperatures, a certain amount of heat added to the system causes a smaller change in entropy than the same amount of heat at a lower temperature. The formula is $\Delta S = \frac{Q}{T}$. The change in entropy is related to heat. Remember that heat is a form of energy transfer, not energy; we can talk about heat only when some change takes place. Please do point out if there is anything wrong with the answers above.
10814
I have a number of measurements of the same quantity (in this case, the speed of sound in a material). Each of these measurements has their own uncertainty.  $$ v_{1} \pm \Delta v_{1} $$ $$ v_{2} \pm \Delta v_{2} $$ $$ v_{3} \pm \Delta v_{3} $$ $$ \vdots $$ $$ v_{N} \pm \Delta v_{N} $$  Since they're measurements of the same quantity, all the values of $v$ are roughly equal. I can, of course, calculate the mean:  $$ v = \frac{\sum_{i=1}^N v_{i}}{N}$$  What would the uncertainty in $v$ be? In the limit that all the $\Delta v_i$ are small, then $\Delta v$ should be the standard deviation of the $v_i$. If the $\Delta v_i$ are large, then $\Delta v$ should be something like $\sqrt{\frac{\sum_i \Delta v_i^2}{N}}$, right?  So what is the formula for combining these uncertainties? I don't think it's the one given in  (though I may be wrong) because it doesn't look like it behaves like I'd expect in the above limits (specifically, if the $\Delta v_i$ are zero then that formula gives $\Delta v = 0$, not the standard deviation of the $v_i$).
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In Arnold's Mathematical Methods of Classical Mechanics, he derives the Hamilton-Jacobi equation (HJE) using a generating function $S_1(Q, q)$ to get  $$ H\left(\frac{\partial S_1(Q, q)}{\partial q}, q, t \right) ~=~ K(Q, t). $$  However, this is different from what I've seen in other physics texts. For example, Goldstein uses the generating function $S_2(q, P, t)$ to get the equation  $$ H\left(\frac{\partial S_2(q, P, t)}{\partial q}, q, t\right) ~=~ - \frac{\partial S_2(q, P, t)}{\partial t}. $$  Why is there this difference? Are the two equations saying the same thing?