I think you just have to expand the wave function to first order in ##\Delta x## and then $$P(\Delta x)=|\Psi(\Delta x)|^2 \Delta x$$
Imagine the area under an infinitesimal interval, in this limit you can substitute multiplication for the integral. I am not entirely sure though.
See the post...
You are not supposed to see those. Are you sure that your browser can render Latex?
I gave an example for you, but if your browser can't render it then there is no point of typing again. Please check that first.
Meanwhile, read about Einstein summation convention. (it seems that this is your...
Fix ##i##, ##j## and then run over ##k##.
For example for ##i=x## and ##j=y## $$\epsilon_{xy,kk}=\epsilon_{xy,xx}+\epsilon_{xy,yy}+\epsilon_{xy,zz}$$
Where a comma denotes a partial derivative. (##A_{,x}=\frac{\partial A}{\partial x}##)
Hope this helps.
I know what you mean. To be honest usually one can solve a problem 'quickly' if he/she has encountered it before, or at least something similar. Otherwise, it would take considerably more time to be solved depending on the complexity of the problem. I don't have much to say except keep...
As already stated by the members above, you can raise and lower it the indices on the stress energy tensor as you like. Why the usual stress tensor ##T^{\alpha \beta}## has two upper (or lower indices)? Maybe because the way they are sometimes defined. For a perfect fluid its defined as...
The expression for momentum for photons in QM is ##p=\hbar k##. Where ##k## is the wave number (##k=\frac{2\pi}{\lambda}##) and ##\hbar## is the Planck constant divided by ##2\pi## . Or equivalently, ##p=\frac{h}{\lambda}## . Your equation for ##p## is for massive particles with non zero rest...
This! I was going to ask why the Riemann tensor is a Tensor although it is made up of Christoffel symbols but then remembered that it is obvious from it's derivation. Or from this definition as well ##[\nabla_i,\nabla_j] \zeta^k=R^k_{~cij}~\zeta^c##. Thanks for your help and also for the references.
Thanks to you, I think that I understand now,. The Ricci scalar ##R=g_{ik}R^{ik}## is a scalar because even if we chose an inertial frame it might not vanish due to the derivatives of Christoffel symbols contained in ##R##, ##\partial \Gamma##. Which means ##R\propto \partial^2g##. So even a...
##G## was calculated to and found to be ##G=g^{ik}( \Gamma^m_{il}\Gamma^l_{km}-\Gamma^l_{ik}\Gamma^m_{lm})##. Isn't this a scalar??
Your explanation makes sense. But I can't relate it to my question. Please refer to the form of ##G## I provided above.
Book: Landau Lifshitz, The Classical Theorey of Fields, chapter 11, section 95.
I have gone through the derivation of Einstein field equations but not without holes to fill and fix in my understanding. Let's start with the action for the grtavitational field ##S_g## which after some explanation...
I have just gone through chapter 14 on the QFT for the gifted amateur by Lancaster and Blundell. Quantising the electromagnetic field results in the Hamiltonian:
$$\hat{H}=\int d^3p \sum^{2}_{\lambda=1} E_p \hat{a}^\dagger_{p\lambda} \hat{a}_{p\lambda}$$
with ##E_p=|p|##. In this post ##p##...
What do you mean by "Classical coordinate system"? If you mean the Cartesian ##(x,y,z)## then it would not work because this is a relativistic situation(Acceleration ##\approx c## !) . So you need Minkowski Coordinates ##(t,x,y,z)##. But since we are dealing with gravitational fields and...
To me this read: The partial derivative with respect to ##\rho## evaluated at ##\rho=\rho_0##.
As in take the derivative first and then set ##\rho=\rho_0##.
I could be wrong though.
If you like doing physics then just do it!
Now, lots of students coming from high schools are shocked when they find out what physics really is. The problem is that they have been watching lots of popular science TV shows or reading nontechnical books. Studying Physics involves doing lots of...
Well, it depends on ##f## and ##g## and not so on the partial derivative. If ##f## and ##g## are "normal" functions like ##f(x)=x^2## for example, then the statement is true. On the other hand, if they represent matrices then generally they wouldn't commute, ie. ##f\cdot g\neq g\cdot f## because...
Nope it is not on the same axis as x. If that was the case then the equation ##x^2+y^2+z^2=\rho^2## wouldn't make sense.
If you are bothered by the provided diagram then just look up another one from another source :wink:
I don't have his book on me but I looked up his notes online here https://preposterousuniverse.com/wp-content/uploads/grnotes-three.pdf (page 56)and the sign is negative.
But check again if you are writing in terms the primed/non-primed coordinates and which you initially started with. I am...
You are absolutely right. The work-energy theorem is true for general forces regardless of them being conservative or not (depends on the resultant force). $$\Sigma W=\Delta KE$$ where $$\Sigma=W_c+W_{nc}$$ Conservative and non conservative, respectively. Is this correct?
It is just a first order differential equation that can be solved by multiplying both sides of the equation by an integrating factor ##U(t)##. Look it up. If you need more help let us know.
If you want to check (or proof) that it works, then just plug the solution ##v## into the differential...
The translation operator translates the quantum state itself by a certain amount.
Say we have the translation operator ##T(a)## that translations the quantum state ##|x\rangle## which gives the eigen value ##x## when acted upon with the operator ##\hat{x}##, by distance ##a## along the ##x##...
Doesn't have to be inside a black hole for that effect to happen. Let's say one is close to the black hole and the other is far away. "Really or perceived" doesn't really make sense. According to relativity it depends on the observer. Type time dilation on Wikipedia.
See above.
Nope. Depends...
As MeJennifer has written, spacetime geometry is rather different past the event horizon. Also we can't tell what is going on inside the horizon using general relativity, loosely put. There is not a mean of communication from inside to outside the event horizon.
Yes, in the absence of fields the work done is equal to the difference of Kinetic energies at two different points: $$W=\Delta KE.$$ Since the velocity is constant, then $$\Delta KE=0.$$
Hi everyone. For anyone who has the book. I am going through Quantum Field Theory for the Gifted by Amateur Tom Lancaster and Stephen J. Blundell. Are the topics enough to prepare me for a course in QFT and then Advanced QFT? Of course I can look for other resources. But I just want to know how...
Hi Faust90
The index ##i## under the sum refers to the subscript ##i## under the ##r##. The other i is the imaginary number. The Kronecker delta gets rid of the exponential and thus the sum on ##i## anyway. Try this website and see if it clears any confusion you are having...
gnnmartin, thank you for your comments. Yes it has been some time since I have asked that question. I did not a have a good foundation or grasp of Tensor calculus and differential geometry which made it hard for me to understand Ray's book which was my first book on GR. I switch to Schutz since...
Thank you both DrGreg and vanhees71. Very easy to follow.
I guess I will read up about elliptic integrals to get a better feel of the integral.
Just one thing, I think that there are easier ways to calculate the deflection. Why did Schutz choose this way to do it? any conceptual/mathematical...
Thanks vanhees71. I have read your post in the thread you have mentioned. tbh I have always thought of a photon in GR textbooks as the classical light rays.
I wish if Schutz stated why he uses some of those substitutions to avoid confusion. What is that second order term doing in (11.52)? Also...
Hi everyone!
I was wondering if we have Hebrew speakers here on PhysicsForums.
I am teaching myself Hebrew and would like to make friends with native speakers or with students of the language.I actually study biblical Hebrew with emphasis on the Yemenite pronunciation, it still useful to have...
Thanks for the reply pervect. Yes you are correct about the definitions. But I actually know what they mean and can do the calculation just fine. I am just not sure why he defined (11.52) in that specific way to simplify (11.49).
Well , when I do Differential Geometry, I think of it as a vector moving on the time axis (dimension). And it can only flow towards the positive direction.
But a vector on the x-axis (imagine a line), can move towards +x or -x and you generalize this to whatever dimension. Time is another type...
Hello Everyone,
I am working through Schutz's A first Course in General relativity. On page 294 he defines the equations (11.52) to simplify equation (11.49) and calculated the deflection of light around the sun. I know that he wants to simplify it and also to preserve the effect of the mass M...
I would show them the postulates and then derive the Lorentz transformations from them. I think they he/she will be satisfied if you showed them how fixing the speed of light leads to the transformation. I could be wrong though.