What happens when you solve for c?

[Joshua514]

In E=mc^2
If you solved for C
you would get c = the plus & minus square root of E divided by M
c = √E/m , -√E/m ?

My question is what would the negative constant be?

 

[NFuller]

My question is what would the negative constant be?

Nothing, speeds (such as c) are magnitudes and thus always positive.

 

[Grinkle]

Look here for a nicely understandable derivation.

https://www.scientificamerican.com/article/significance-e-mc-2-means/

I don't see any physical significance to the negative root from the above derivation.

 

[Joshua514]

I think we're missing half of the equation here, if light is matter than theoretically if negative magnitudes existed wouldn't that be dark matter?

 

[NFuller]

if light is matter

Light is electromagnetic radiation. Even though light is sometimes described using photons, you should not be thinking of light as little balls of material.

if negative magnitudes existed

The problem is that they don't.

 

[pervect]

I'd generally interpret the sign of a velocity as being it's direction. Speed is defined as the magnitude of the velocity, so it has no sigh as others have mentioned. In terms of velocity, which can have a sign, +c is an object moving to the right, -c is an object moving to the left. Or vica versa, depending on how you chose your coordinates.

 

[Dale]

I think we're missing half of the equation here, if light is matter than theoretically if negative magnitudes existed wouldn't that be dark matter?

Do you have any scientific reference which associates dark matter with -c? I strongly doubt it.

 

[Ibix]

I think negative c in this context just means you've flipped a sign convention or two, doesn't it? At least if you are going to be consistent, you should argue that the modulus of the four-velocity be $(-c)^2$ too, which implies that you've flipped the sign on your time coordinate, I think.

In other words you need to glue a minus sign to the displays of all your clocks. That's all. It doesn't actually make any difference to anything.

 

[ChrisVer]

if light is matter

that's what devalues everything coming after it... (that if evaluates to "FALSE" so someone wouldn't have to keep reading)
Light is not matter, and it doesn't follow the same equation of state. As a result it can't be associated with DM or DE.

I think we're missing half of the equation here

Not really, we don't.
First of all, $E = mc^2$ is only true for massive particles that can't travel at the speed of light. The speed of light in this case is just a conventional constant (a natural choice) that makes the units right. This equation can also be seen as $E= m$ given that masses are given in units of energy... and a more complete equation is $E^2=m^2-p^2$ as in your case $m$ is the relativistic mass (an outdated quantity) or the rest mass (if the massive object is at rest; so in a special reference frame).

 

[Mister T]

In E=mc^2
If you solved for C
you would get c = the plus & minus square root of E divided by M

If you solved $A=\pi r^2$ for $r$ you'd get $\pm \sqrt{\frac{A}{\pi}}$.

Does that mean you could have a circle with a negative radius?

 

[FactChecker]

When you take the square root of a positive number, you have two possible solutions that you need to consider. That does not mean that they both make sense. You need to consider each possibility and see if one or both could be a valid answer. In this case, the negative answer does not make physical sense.

 

[Vanadium 50]

I think we're missing half of the equation here, if light is matter than theoretically if negative magnitudes existed wouldn't that be dark matter?

ANd now we are in personal theory land.