Recent content by fdsa1234

Ah, so ##E = \frac{mc^2}{\sqrt{1 - v^2/c^2}}## simply becomes ##E = \frac{m}{\sqrt{1 - V^2}}##. Now I see it. Thank you.

I already know the solution to this problem, but I'm not sure exactly why it works out the way it does, so I'm looking for an explanation. Homework Statement A particle accelerator accelerates electrons at 40 GeV in a pipe 2 miles (3218.69 metres) long, but only a few cm wide. How long is the...

Thank you so much. For part b), can I just plug numbers into a Lorentz equation, say, ##t' = γ(t - vx/c^2)##? And I'm assuming the resulting speed should be a significant fraction of c?

So, ##(3)^2 = -c^2(10^-8)^2 + \Delta x'^2## I'm getting ##x' = 4.2425.## Please tell me that's correct.

That's what I thought. Is ##\Delta s'## also equal to 3 because it is invariant?

Should it not also equal 3m? ##\Delta s^2 = -c^2(0)^2 + (3)^2##? ##\Delta s'^2=-c^2(10^-8)^2 + \Delta x'^2## But how can I solve that if both ##\Delta s'## and ##\Delta x'## are unknown?

I see. I hadn't thought to use that; the context it was used in the lecture was for four-vectors with multiple spatial dimensions. So, ##\Delta s^2=-c^2(10^-8)^2 + (3)^2## (that should be 10^-8, but it won't show up properly)? From that I get ##\Delta s^2 = (-1) * (3 * 10^8)^2 * (10^-8)^2 +...

The invariant interval ##I = -c^2 * delta(t)^2 + d^2##?

3 +/- 3, that is. I think it's +3? So, the new separation is 6 metres? Or am I completely off here...? Apologies for the double post; I exceeded the post edit time limit.

Hmm... ct? I thought that the answer might be 3 +/- 1 metre (one or the other, not both), but I'm not sure...

The speed of light?

1. Two events occur simultaneously in an inertial reference frame, separated by a distance of 3 metres. Within a different inertial frame that is moving with respect to the first, one event occurs 10^-8 seconds later than the other. (a) In the moving frame, what is the spatial distance between...

I see. Thanks a lot for your help, guys.

Yeah, I tried that equation after I posted and got 8.33... s, which I believe must be correct.

Eh... V2 = V1 + at?