Lorentz Transformation: Solving Homework Statement

yellowputty
Messages
8
Reaction score
0

Homework Statement



Two events occur at the same place in an inertial reference fram S, but are separated in time by 3 seconds. In a different frame S', they are separated in time by 4 seconds.

(a) What is the distance between the two events as measured in S'?
(b) What is the speed of S relative to S'?

Homework Equations



I'm presuming:

t' = gamma*(t-ux/c^2)

The Attempt at a Solution



I have the answer, and a hint saying to use the interval S^2, but I have no idea what that means, and where I start. When I look and the relevant Lorentz equations, they involve velocity, and I do not have a velocity here.

Could you please point me in the right direction?

Thank you in advance.
 
Physics news on Phys.org
The hint is implying that you use the invariant space-time interval to solve this problem. The following quantity is called the space time interval:

(\Delta s)^2= (\Delta x)^2 + (\Delta y)^2 + (\Delta z^2) - (c\Delta t)^2

This quantity is a Lorentz scalar and is thus invariant over Lorentz transformations (it is the same in all inertial frames). So, in one dimension this means:

(\Delta x)^2- (c\Delta t)^2=(\Delta x')^2- (c\Delta t')^2

Can you use this the solve the problem?
 
G01 said:
The hint is implying that you use the invariant space-time interval to solve this problem. The following quantity is called the space time interval:

(\Delta s)^2= (\Delta x)^2 + (\Delta y)^2 + (\Delta z^2) - (c\Delta t)^2

This quantity is a Lorentz scalar and is thus invariant over Lorentz transformations (it is the same in all inertial frames). So, in one dimension this means:

(\Delta x)^2- (c\Delta t)^2=(\Delta x')^2- (c\Delta t')^2

Can you use this the solve the problem?

Do I find (\Delta t) by doing SQRT[(4^2)-(3^2)] = ROOT 7

Then at they are both at the same coordinates in the inertial reference frame, we can ignore x , y and z. Therefor S equals the root of (c^2)*(ROOT 7) = 7.9x10^8m

Is this correct?
 
yellowputty said:
Do I find (\Delta t) by doing SQRT[(4^2)-(3^2)] = ROOT 7

Then at they are both at the same coordinates in the inertial reference frame, we can ignore x , y and z. Therefor S equals the root of (c^2)*(ROOT 7) = 7.9x10^8m

Is this correct?

You should only have one factor of c in your final line, since you take the square root of c^2 when solving for the answer.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top