Timeline for Why can light (photons) bends in a curve through space without mass?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 4, 2020 at 16:03 | history | edited | CommunityBot |
Commonmark migration
|
|
| Jul 19, 2018 at 20:37 | review | Suggested edits | |||
| Jul 19, 2018 at 21:31 | |||||
| Apr 12, 2014 at 6:30 | comment | added | Ruslan | @JanHudec if you use relativistic mass to rewrite Newtonian equations of motion, then you'll have to deal with dependency not only on velocity, but also on angle between force exerted on a particle and its velocity. I.e. you have anisotropic mass. In this case equations will become even more complicated and will include $m_{||}$ and $m_\perp$ (details). Why would you do it?Just to simplify the most trivial equations? This simplification misleads. Better just use correct relativistic machinery and drop the notion of relativistic mass. | |
| Apr 11, 2014 at 21:48 | comment | added | Cole Tobin | @JanHudec However, energy is not mass. They can be converted into each other a la $E^2 = m_0^2 c^4 + p^2 c^2$, but they aren't the same. Wikipedia fun. | |
| Apr 11, 2014 at 21:14 | comment | added | Jan Hudec | @Ruslan: Since relativistic mass is equivalent to energy anyway I understand it's easier to rewrite the formulas for energy instead, but as long as we are attempting to reuse formulas from Newtonian dynamics here, relativistic mass is needed in them. | |
| Apr 11, 2014 at 20:56 | comment | added | Ruslan | @ChocoPouce main idea is not that space-time is curved by mass. It's curved by energy, and photons do have energy, that's why they can curve space so that, for example, two initially parallelly moving photons' paths would intersect (if we don't consider quantum uncertainty). | |
| Apr 11, 2014 at 20:54 | comment | added | Ruslan | @JanHudec no they don't. Modern physicists try to no longer use the notion of relativistic (i.e. non-rest) mass. See this for details. | |
| Apr 11, 2014 at 20:39 | comment | added | Jan Hudec | @ColeJohnson: Fortunately, photons do have mass (equal to $\frac{h\nu}{c^2}$), so you don't get that. They just don't have rest mass. | |
| Apr 11, 2014 at 20:38 | comment | added | Jan Hudec | @ColeJohnson: Yes, if you plug in $0$ for $m_1$, you get $F_g = G\frac{0 m_2}{r^2}$, but you also get $a_g = \frac{\frac{0 m_2}{r^2}}{0}$. | |
| S Apr 11, 2014 at 19:50 | history | suggested | Reinstate Monica -- notmaynard | CC BY-SA 3.0 |
Minor grammar, spelling, punctuation corrections
|
| Apr 11, 2014 at 19:44 | review | Suggested edits | |||
| S Apr 11, 2014 at 19:50 | |||||
| Apr 11, 2014 at 19:15 | comment | added | Cole Tobin | You forgot that photons have no mass in your list at the top. Because if you plug in 0 for, say $m_1$, you get $F_g = G \frac{0 \times m_2}{r^2}$ which simplifies to $F_g = 0$. You did mention it in your summary, but I feel it should be in the top list also. | |
| Apr 11, 2014 at 14:57 | vote | accept | Poomrokc The 3years | ||
| Apr 11, 2014 at 14:54 | history | answered | ChocoPouce | CC BY-SA 3.0 |