I would assume that $\Delta m$ would mean a finite change in mass and $dm$ to be an infinitesimal change in mass. But many physics textbooks use it to denote a small but finite mass and an infinitesimal mass element respectively while writing a differential equation. Why? And how does this make sense?  Similarly, $dU$, which should mean the infinitesimal change in the potential energy function is treated as the "small" potential energy of mass $dm$. Same for $dF$ etc.  One e.g would be to look at the derivation of the gravitational potential energy of a point mass and a shell. The potential energy of the ring is changed from $U_i$ to $dU$ and the mass of the ring is taken as $dM$. Doesn't $d$ mean an infinitesimal change in some quantity?
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?
89997
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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I'm a web developer and I have to change an online course. The course teaches Advanced Nuclear Theory. In the 'Mass Defect and Binding Energy' chapter, it has this formula:  $$\Delta M = Z(m_p) + N(m_n) - \frac{A}{Z}(M).$$  In the formula above, the letter 'p' and 'n' are written in subscript, and for the 'AZM' part, the letter A (written in superscript) is above the letter Z (which is in subscript). Above this equation, this is written  1/1(p) + 1/0(n) -&gt; 2/1(D)  mass of proton = 1.007272 u mass of neutron = 1.008665 u mass of deuterium nuclear = 2.013553 u   . Below the equation which I wrote above (delta M = Z(mp) + N(mn) - A/Z(M)), the worker / user is asked to fill in the blank:  ∆M = __ x __ + __ x __ - __  B.E. of nucleaus = 931.5 x __ MeV  When the neutron and proton combine, the nuclear force pulling them together does __ MeV of work.   Now, can anyone tell me what the correct answers are to the blanks? I'm just a web developer and I have to write a program which checks that the user filled in the correct answers, but I do not know the correct answers. I have some knowledge of physics and I'm guessing mp = mass of proton, mn = mass of neutron and then M = mass of deuterium nucleus, but what about the other blanks?
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I'm currently studying the Dirac delta function using a textbook which unfortunately provides only partial solutions to its explanations. Why does $\lim\limits_{\epsilon\rightarrow 0}\int_{-\infty}^{\infty}f_\epsilon(x)g(x)dx=g(0)$ if $ f_\epsilon=\frac{1}{2\epsilon}e^{-\frac{|x|}{\epsilon}}$? Any help is very much appreciated.
120640
I've stumbled across the following notation a couple times reading physics articles on wikipedia: $$\partial_{\mu}$$ But what does it mean? They don't clarify.  Source:
111636
My understanding of Quantum physics and String Theory is very basic and I don't yet have a grasp on the maths, but in my research I have come up with a question.  Does a decrease in temperature also signify/create a decrease in mass?  I am quite willing to believe the concepts I think I am beginning to understand are totally misconceptions on my part.  But, as I see it mass is a product/mark of disruption to the Higgs field as a particle moves.  If these particles are ultimately composed of vibrating Strings, their mass is given by the rate at which they vibrate and how it interacts with the Higgs field.  I don't have any idea if, as the temperature (as we know it) approaches 0k, the string/quark/sparticle would change its state in any way or if the changing of the state of the String would change what type of particle it represents.  It seems I'm more full of questions than answers, but that's why I love these topics!  I know what I'm learning is new if the best and brightest are still working on the 'basics' to just make their systems work :).
37875
Why is the following equation true? $$\frac{\partial \mathbf{v}_i}{\partial \dot{q}_j} = \frac{\partial \mathbf{r}_i}{\partial q_j}$$ where $\mathbf{v}_i$ is velocity, $\mathbf{r}_i$ is the displacement, and $q_j$ is the generalized coordinate into which $\mathbf{r}_i$ is transformed. In reading further, I find that it's related to $$ \mathbf{v}_i \equiv \frac{\mathrm{d}\mathbf{r}_i}{\mathrm{d}t} = \sum_k \frac{\partial \mathbf{r}_i}{\partial q_k}\dot{q}_k + \frac{\partial \mathbf{r}_i}{\partial t} $$ I know that in transforming the virtual displacement $\delta\mathbf{r}_i$ into generalized coordinates, I can use $$ \delta\mathbf{r}_i = \sum_k \frac{\partial\mathbf{r}_i}{\partial q_k}\dot{q}_k $$ The first equation above is of course equivalent to $$\frac{\partial \dot{\mathbf{r}}_i}{\partial \dot{q}_j} = \frac{\partial \mathbf{r}_i}{\partial q_j}$$ I'm not sure why the dots vanish just like that. How do these all connect? In a non-mathematical explanation, I can understand that as $q_j$ changes, $\mathbf{r}_i$ changes as well. In the same way, as $\dot{q}_j$ changes, $\mathbf{v}_i$ changes, too. I'd like to know, mathematically, how these changes (for $\mathbf{r}_i$ and $\mathbf{v}_i$) turn out to be equal.
66015
Sorry if this is really naive, but we learned in Newtonian physics that the total energy of a system is only defined up to an additive constant, since you can always add a constant to the potential energy function without changing the equation of motion (since force is negative the gradient of the potential energy).   Then in Quantum Mechanics we showed how the ground state of a system with potential energy $V(x) = \frac{1}{2} m \omega^{2} x^{2}$ has an energy $E_{0}=\frac{1}{2} \hbar \omega $.   But if we add a constant to $V(x)$ won't that just shift the ground state energy by the same constant? So in what sense can we actually say that the ground state energy has an absolute value (as opposed to just a relative value)? Is there some way to measure it?   I ask this in part because I have heard that Dark Energy might be the ground state energy of quantum fields, but if this energy is only defined up to a constant, how can we say what it's value is?
31834
I don't understand why it is said that the (rest) mass of a system is not conserved in relativity. I mean, the momentum of a system is conserved (i.e.: it remains constant in a frame of reference without any external influence). Also the energy of a system is conserved in relativity: it doesn't change without any external agency  and the (rest) mass of a system is just  $$m^2=E^2-p^2$$  $E$ is constant, $p$ is constant, how can $m$ change?  For example, entropy is not conserved in a system, that means that the entropy of the system will increase spontaneously with time, which is really the case. But, is this true for mass?   here is the problem from Griffiths and exact solution as given in the Griffiths,  Two lumps of clay, each of rest mass $m$, collide head-on at $3c/5$ and they stick together. Question: what is the rest mass $M$ of the composite lump?  Solution: In this case conservation of momentum is trivial it is zero before and zero after. The energy of each lump prior to the collision is $$mc^2/\sqrt{1-v^2/c^2} =5mc^2/4.$$ The energy of the composite lump after the collision is $Mc^2$ (since its at rest). so the conservation of energy says: $5mc^2/4 + 5mc^2/4 =Mc^2$ and hence $M=5m/2$.  Notice that this is greater than the sum of the initial masses! Mass was not conserved in this collision; kinetic energy was converted into rest energy, so the rest mass increased.
42608
Is $0 \, \mathrm m = 0 \, \mathrm s = 0 \,\mathrm {kg} = 0$? How do we define $[0 \, \mathrm m]$?  I once was given an assignment where I was asked to deduce and write down some physical quantity. It turned out that this quantity was $0$ in one unit or another, so I decided to drop the physical units because I figured that it didn't matter anyhow.  But I got to thinking:  If $0 \, \mathrm m = 0$, then we could add units of different dimensions (an operation not normally defined) like so: $5 + 0 \,\mathrm m = 5$.   Does $[ \cdot ]$ map $0 \, \mathrm m$ to both $0$ and $L$? Is there an established regime of thinking about physical quantities as an abstract algebra? Perhaps as a vector space where formal sums such as $5 \, \mathrm m + 10 \, \mathrm s$ are allowed? Would it be improper to drop the units of measurement of $0 \,\mathrm m$ like this in an academic paper?
61836
I've read many question/answer threads here on SE re: justification for the algebraic manipulation of $\mathrm{d}y/\mathrm{d}x$ in the standard formulation of calculus. I worked up my own shot at a justification using the definition of the derivative and would like to get the input of others whether it holds. It doesn't involve pushforward maps, non-standard analysis (two things I don't yet know), or any other tool than simple limit algebra/laws, the definition of the derivative, and (implicitly), the chain rule that makes the substitution possible. If it does hold as a justification as far as it goes, i.e., in one-dimension, it would seem to me, at least, to be a simple and clear justification in the form of an example that could be useful to other beginners. If goes wrong somewhere please let me know for my own understanding.   Integrate $\int \frac{x}{1+x^2}\mathrm{d}x$.  We'll do this by substitution. We want to restate the original problem in terms of $u$. That means that we want to state $x$, $1+x^2$, and $\mathrm{d}x$ in terms of $u$. Let $u=1+x^2 \rightarrow \frac{\mathrm{d}u}{\mathrm{d}x}=2x \rightarrow \mathrm{d}x=\frac{\mathrm{d}u}{2x}$. Let's justify that last step as that's where the justification (that I speak of) is required:  By definition, $\frac{\mathrm{d}u}{\mathrm{d}x}=2x$ is defined as $$ \lim_{\Delta x \to 0}\left( \frac{u(x+\Delta x)-u(x)}{\Delta x} \right)=2x $$ Recall that the limit of a quotient is equal to the quotient of the limits as long as the denominator is not equal to zero, thus the foregoing becomes $$ \frac{\lim_{\Delta x \to 0}\left( u(x+\Delta x)-u(x) \right)}{\lim_{\Delta x \to 0} \Delta x } =2x $$ Multiplying both sides by $\lim_{\Delta x \to 0}\Delta x$ and then dividing both sides by $2x$ yields $$ \frac{\lim_{\Delta x \to 0}\left( u(x+\Delta x)-u(x) \right)}{2x} =\lim_{\Delta x \to 0} \Delta x $$ which, in the limit, is equal to $$ \frac{\mathrm{d}u}{2x}=\mathrm{d}x $$ Rearranging we have $$ dx=\frac{\mathrm{d}u}{2x} $$ as desired.
86278
I'm used to people using the symbol $\equiv$ when defining quantities. I might see expressions like  $$\frac{df}{dx} \equiv \lim_{\delta\to 0} \frac{f(x+\delta) - f(x)}{\delta} = 2x \,.$$  where the first part is what is meant by the symbol $df/dx$, whilst the second is an equality that holds in the specific case being considered. On the other hand, what I understand the symbol $\equiv$ to actually mean is 'equivalent to'. A good example of such a usage would be  $$ \sin^2 \theta + \cos^2 \theta \equiv 1 \,. $$  The two usages to me seem rather different. The statement above is not a definition of $\sin$ and $\cos$, and certainly not a definition of $1$! Rather it is a statement with content. The first usage of $\equiv$ that I gave is, on the other hand, content-less, for it only serves to define a symbol.  So what I'm wondering is this:  Are the two meanings I've outlined above in fact not all that different? Or if they are, why is it common to use the same symbol for both? Do we need a new symbol for 'defined to be'? I believe I have seen before the symbol  $$ \stackrel{\mathrm{def}}{=} $$  but it is used very rarely. It is not possible that the different usages of the symbol could cause confusion among students? Or would most people agree that meaning is clear from context?  Thank you.
74191
This was supposed to be a long question but something went wrong and everything I typed was lost. Here goes.   Why is  $k = \dfrac{1}{4\pi\epsilon_0}$ in ?  Is this an experimental fact?  If not, what is the significance of this definition?
15566
This suggests, as a shortcut notation, the concept of lowering indices; from any vector we can construct a (0, 1) tensor defined by contraction with the metric: $$A_\nu ≡ g_{\mu\nu}A^\mu$$  so that the dot product becomes $$g_{\mu\nu}A^\mu B^\nu = A_\nu B^\nu$$  We also define the inverse metric $g^{\mu\nu}$ as the matrix inverse of the metric tensor: $$g^{\mu\nu}g_{\nu\rho} = δ^\mu_\rho$$  where $δ^\mu_\rho$ is the (spacetime) Kronecker delta. (Convince yourself that this expression really does correspond to matrix multiplication.) Then we have the ability to raise indices: $$A^\mu = g^{\mu\nu}A_\nu$$ Note that raising an index on the metric yields the Kronecker delta, so we have $$g^{\mu\nu}g_{\mu\nu} = δ^\mu_\mu = 4$$  I don't understand why $δ^\mu_\mu = 4$. Shouldn't it equal 1?
90395
We know that if an object has been lifted a distance $h$ from the ground then it has a potential energy change:  $$\Delta U = mgh $$  so $h$ is proportional to $\Delta U$.  However, we have also the gravitational potential energy law:  $$ U= -\frac{G M m}{r} $$  where the distance is inversely proportional to the potential energy.  What did I miss? Is the distance of the object proportional or inversely proportional to the potential energy?
44071
This is pressure in Newtonian mechanics:   $$P=\frac {dF}{dA}.$$ What does this mean? (Doesn't it mean that force is a function of area?) What type of function is force?
78483
I am currently reading through a document  from MIT, page 4, and I am a little confused as to one of the concepts that they use.  In this document, there is mention of a . Could someone possibly define this for me please? I can't find anything too clear on the Internet.  Is this synonymous with the first moment of mass?