Hello,
First of all, I’d like to tell you that I’m not fluent in English (I think it’s important to say) and that’s a reason why I’m asking my questions here. Sorry for all the mistakes in this text.
When I was in my bedroom, looking at the light bulbs of the ceiling, I wondered about the...
Thanks to you for your answers,
Well, I understand now why I was confused, even if I'm not quite sure how the asteroid "gains" its kinetic energy. Probably I should wait a bit more to have a "richer" background :smile: .
Yes, I haven't enough knowledge to expect to understand what SR and GR...
Hello everybody,
Let’s say a rocket which is at rest relatively to an asteroid. The rocket engine start and the rocket is launched toward the asteroid’s neighborhood. When the rocket engine is working, the rocket accelerates. Few minutes later, the rocket engine turns off and the rocket is now...
Hello,
In fact, energy doesn't equal mass, but according to this well known formula, E=mc² indicates that whatever are the speed or the motion of a body, if it has a mass, it has energy.
E=mc² comes from a bigger formula which is E^2 = p^{2}c^{2}+m^{2}c^{4}. Using that equation, you can...
Hi,
I'm not sure if I'm at the right place to write this message, so please excuse me if I'm not.
I have read a bit about tidal forces, but I still don’t understand what they are, how they act on different body masses, etc. So, I thought I could find answers here. I would greatly appreciate it...
Hello,
Pete, there is something I would like to understand at the number (7) of your website :
v\gamma t' = v\gamma^2 t - \gamma^2 \beta^2 x = v\gamma^2 t - \gamma^2 vx/c^2
Why does \beta^2 x = vx/c^2 if \beta^2 = v^2 / c^2?
Isn't v\gamma t' = v\gamma^2 t - \gamma^2 v^2 x/c^2 = v\gamma^2...
Hello,
Thanks for the answers. In my first post, I wanted to help, but I realize the good answer is much more difficult than I thought :shy: Thanks again.
Zeit
Hello,
I don't understand why I should be wrong. Rest mass, as its name suggests me, cannot increase : it is an invariant mass. The "relativistic mass" is, as I understand it, a wrong concept ; we should talk of "relativistic momentum". So, may be is it better to use E² = (mc²)²+(pc)² than E...
The equation E = mc² comes from a larger : E² = (mc²)² + (pc)²
When the studied object is motionless, p = 0, so E² = (mc²)². So, even when an object is motionless, it has an energy. So, we can say that mass is a kind of energy.
Sorry, but what myoho.renge.kyo quotes is true. As I said, if you do not directly apply this result to another reference frame (in which case we use Galileo's transformations, so classical mechanics). But, the fact is that Galileo was right... for an only reference frame...
Look at my example...
Hello,
Imagine that you are near a road. There are two cars, a blue one (B) and a red one (R). Note the length of R as L. The experience is to mesure the time needed by B (which always travels at a constant speed) to traverse L.
First, imagine that R is motionless. At t=0, the front of B...