The speed of light varies across the universe...a theory

[Jalopy]

e=mc²

c=√e/m

mass is how heavy an object is. Meaning c, the speed of light, depends to some extent on the force of gravity. 

And gravity is different on different planets. 

So logically, the speed of light is...different from planet to planet? 

Einstein himself said so when he said e=mc²

[exchemist]

24 minutes ago, Jalopy said:

e=mc²

c=√e/m

mass is how heavy an object is. Meaning c, the speed of light, depends to some extent on the force of gravity. 

And gravity is different on different planets. 

So logically, the speed of light is...different from planet to planet? 

Einstein himself said so when he said e=mc²

It hardly needs saying on a forum like this one, but your post is idiotic.

Your "e" is "E", meaning (rest) energy. E is proportional to m, c being constant.  Furthermore m is mass, which is not how heavy something is and thus does not depend on the force of gravity.

 

Edited March 3, 2022 by exchemist

[swansont]

c is invariant, which means it's the same in all inertial reference frames. If you are at a fixed location in a gravitational field, you are in an accelerated frame, not an inertial one. You will locally measure c to be the same, but that might not be the case if you measure it in reference to some other frame.

The difference, though, might be negligible with regard to the precision of a measurement you are making.

[MigL]

If you attempt a meaningful post on a science forum, you should, at least, know the difference between mass and weight.
If you don'twish to continue being ignorant of such matters, maybe you should ask.
Might save you some more embarassement.

[Jalopy]

1 hour ago, exchemist said:

It hardly needs saying on a forum like this one, but your post is idiotic.

Your "e" is "E", meaning (rest) energy. E is proportional to m, c being constant.  Furthermore m is mass, which is not how heavy something is and thus does not depend on the force of gravity.

 

even so,

gravity =weight/mass 

therefore

c=√e/m =√e/(gravity/weight)

so c varies if gravity varies. 

c on earth is not the same as c on venus. 

c in outer space is.....null 

wow. 

That would explain why outer space is so dark.

1 hour ago, swansont said:

c is invariant, which means it's the same in all inertial reference frames. If you are at a fixed location in a gravitational field, you are in an accelerated frame, not an inertial one. You will locally measure c to be the same, but that might not be the case if you measure it in reference to some other frame.

The difference, though, might be negligible with regard to the precision of a measurement you are making.

please explain how c can be constant everywhere in the universe, given that;

mass=gravity/weight

therefore

c=√e/(gravity/weight)

so c varies if gravity varies. 

Edited March 3, 2022 by Jalopy

[exchemist]

6 minutes ago, Jalopy said:

even so,

gravity =weight/mass 

therefore

c=√e/m =√e/(gravity/weight)

so c varies if gravity varies. 

c on earth is not the same as c on venus. 

c in outer space is.....null 

wow. 

That would explain why outer space is so dark.

please explain how c can be constant everywhere in the universe, given that;

mass=gravity/weight

therefore

c=√e/(gravity/weight)

so c varies if gravity varies. 

So you simply dismiss my correction of your wrong statement by saying "even so"and just repeating the silly error.  

This is a waste of time. 

[swansont]

10 minutes ago, Jalopy said:

please explain how c can be constant everywhere in the universe

I don't know how you would expect me to do that, when I have told you that it's not, and conditions under which it fails to hold.

[Jalopy]

2 minutes ago, exchemist said:

So you simply dismiss my correction of your wrong statement by saying "even so"and just repeating the silly error.  

This is a waste of time. 

mass=gravity/weight

therefore

c=√e/(gravity/weight)

so c varies if gravity varies. 

in outer space, gravity = 0, therefore c = infinity.

true or false?

Edited March 3, 2022 by Jalopy

[exchemist]

1 minute ago, Jalopy said:

I rescind my 'even so' and venture the following response

please explain how c can be constant everywhere in the universe, given that;

mass=gravity/weight

therefore

c=√e/(gravity/weight)

so c varies if gravity varies. 

...[click]....

[Jalopy]

1 minute ago, exchemist said:

...[click]....

that doesnt explain anything!

4 minutes ago, swansont said:

I don't know how you would expect me to do that, when I have told you that it's not, and conditions under which it fails to hold.

is this conclusion correct, that

since

c=√e/(gravity/weight)

in outer space, gravity = 0,

therefore c = infinity in outer space.

 

[swansont]

5 minutes ago, Jalopy said:

 

is this conclusion correct, that

since

c=√e/(gravity/weight)

in outer space, gravity = 0,

therefore c = infinity in outer space.

 

No, since mass=gravity/weight is nonsense, as I have explained in another thread. Nothing valid can be derived from it.

[Jalopy]

15 minutes ago, swansont said:

No, since mass=gravity/weight is nonsense, as I have explained in another thread. Nothing valid can be derived from it.

I corrected that typo! Here's my fresh theory

m= w/g

therefore 

c=√e/(w/g)

in outer space, g = 0

therefore

c=√e/(w/0)

c=√e/infinity

c=0 

That means, if a light beam was passed through outer space, it would stop moving and yet exist, for energy cannot be created or destroyed.

Then if gravity were thrust upon it, it would start moving, but slowly. 

That would explain how Einstein actually took a ride on a light beam, that's how he did it!  It was a slow moving light beam 

 

Edited March 3, 2022 by Jalopy

[swansont]

Dividing by zero is undefined, which is what happens when you try and use a crappy definition, which is what W/g is.